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You are lying on your back on the slope of a grassy hill, watching the clouds go by overhead. Which is the best statement about the force of friction exerted on you by the ground in this situation?

A force of static friction pointing "up the hill", parallel to the surface of the hill, is applied to you by the ground.

(d) What will the acceleration be if the force is doubled and the object's mass is halved?

Finally, if we double the force and halve the mass, we get 2F F 2 2 apartd=m/2=4 m =4a =47.50m/s =30.0m/s Therefore, if we double the force and halve the mass, we will get four times the acceleration, equal to 30.0 m/s2.

A 6.00 kg block is in contact with a 4.00 kg block on a horizontal frictionless surface as shown in the figure. The 6.00 kg block is being pushed by a horizontal 20.0 N force as shown. What is the magnitude of the force that the 6.00 kg block exerts on the 4.00 kg block?

I've drawn a pictorial representation of the situation on the extreme right above, and two separate free-body diagrams for the 6 kg and 4 kg blocks respectively on the left and middle above. All forces are marked with symbols that identify uniquely whether they are acting on the 4 kg block or the 6 kg block (e.g., normal force on the 4 kg block is n4, force of gravity on the 6 kg block is Fg6, etc.). The force of the 6 kg block on the 4 kg block, marked as F6on4 in the free-body diagram for the 4 kg block, is an action-reaction pair with the force of the 4 kg block on the 6 kg block, marked as F4on6 in the free-body diagram for the 6 kg block. First, it is important to realize that the system of two blocks is undergoing an acceleration to the right. Why? The surface is frictionless, and the system of (6+4) kg is experiencing a net force of 20.0 N to the right. By Newton's second law, it must be accelerating with a= 20.0N =2.00m/s2 (6+4) kg Therefore, each block, the 6 kg as well as the 4 kg, is being accelerated to the right by 2.00 m/s2. Now, consider the 4 kg block, for which I've drawn the free-body diagram in the middle panel of the figure above. The free-body diagram for the 4 kg block reveals only one force acting to the right, the force of the 6 kg block on the 4 kg block, which I've named F6on4. Since this is the net force on the 4 kg block, the 2.00 m/s2 acceleration of the 4 kg block is being produced by this force. Therefore, we can write from Newton's second law (ΣF = ma) that F6on4 = 4.00 kg 2.00 m/s2 = 8.00 N Therefore, the 6.00 kg block exerts a force of 8.00 N on the 4.00 kg block. Alternatively, you could use the free-body diagram to find F4on6, since that will have the same magnitude as F6on4. From the horizontal direction in the free-body diagram for the 6 kg block, we can write from Newton's second law (ΣF = ma) that 20.0 N − F4on6 = 6.00 kg 2.00 m/s2 so that matching the answer calculated from the 4 kg block. F4on6 = 20.0 N − 6.00 kg 2.00 m/s2 = 20N − 12N = 8.00 N

The sled dog in the figure below drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10. If the tension in rope 1 is 200 N , what is the tension in rope 2?

I've drawn two free-body diagrams above, one for sled A (extreme left) and the other for sled B (in the middle, i.e., to the left of the physical picture on the right). In each, we have the usual force of gravity on the sled acting downward (FgA and FgB respectively), the normal acting perpendicular to the surface and upward (nA and nB respectively), and kinetic friction to the left, opposite to the direction of motion (fk1 and fk2 respectively). Rope 1 is pulling on both A and B, so tension T1 is in both free-body diagrams; T1 points to the right in the free-body diagram for A and to the left in the free-body diagram for B. Rope 2 is pulling to the right on B, so it is only in the free-body diagram for B, pointing to the right. First, we need to check what kind of motion the system is undergoing. We have FgA = (100 kg)(9.80 m/s2) = 980 N and from equilibrium in the vertical direction, we get nA =FgA =980N Then, since the coefficient of friction between the sleds and the snow is μk = 0.10, we get fk1 =μknA =(0.10)980N=98N From the free-body diagram for sled A, we see that T1 = 200 N is acting to the right, and we have just calculated above that fk1 = 98 N is acting to the left. Thus, sled A is subject to a net force toward the positive x-direction, and it must be accelerating to the right. From Newton's second law, we can write the net force (T1 − fk1) equal to the mass times the acceleration: T1 − fk1 = (100 kg) a nA T1 fk1 T1 FgA nB fk2 F T2 gB so that Assuming that the ropes remain taut, sled B must also be undergoing the same acceleration of 1.02 m/s2. From the free-body diagram for B, we have T2 − T1 − fk2 = (80 kg) (1.02 m/s2) a=T1−fk1 =200N−98N=1.02m/s2 100 kg 100 kg so that Given that T1 = 200 N, and fk2 = μknB = 0.10(80 kg) (9.8 m/s2) = 78.4 N, we get T2 =(80kg)(1.02m/s2)+200N+78.4N=360N Therefore, the tension in rope 2 is 360 N.

A 6.0 kg box slides down an inclined plane that makes an angle of 39◦ with the horizontal. If the coefficient of kinetic friction is 0.12, at what rate does the box accelerate down the slope?

In the figure below on the right, I've shown the box on the inclined plane. I've also marked with dashed lines the x and y axes that we'll be working with in this problem; it is convenient to choose the x axis along the inclined plane and the y axis perpendicular to the plane. n fk Fg sin θ θ Fg y x θ Fg cos θ θ In the figure above on the left, I've drawn a free-body diagram showing the forces acting on the box. In this figure, I've kept the x and y axes inclined just as in the figure on the right. There are three forces in this problem: the force of gravity, Fg, acting downward, the normal n acting perpendicular to the inclined plane, and the force of kinetic friction fk, acting opposite the direction of motion and hence up the slope of the inclined plane. Since n and fk point along our chosen axes, the only force that needs to have its components determined is Fg. You should look carefully at the geometry of the figure and convince yourself that if θ is the angle of the inclined plane, then the angle made by Fg with the negative direction of the y-axis is also equal to θ. Learn this once and know it for all time (especially so you don't have to work it out every time). The components of Fg are then Fg cos θ along the negative direction of the y-axis and Fg sin θ along the negative x-axis (see the figure above). From the free-body diagram on the left above, it is clear that the net force down the slope is then (Fg sin θ − fk), and dividing this net force by the 6.0 kg mass of the box will give us the rate at which the box accelerates down the slope. It is easy to find Fg by doing Fg = mg = (6.0 kg)(9.80 m/s2) = 58.8 N Next, we can find the force of kinetic friction from fk = μkn, and from equilibrium along the y-direction (see the figure above for the y-direction), we get that n = Fg cos θ fk = μkn = μk(Fg cos θ) = 0.12(58.8 N) cos 39◦ = 5.4835 N so that From the figure above on the left, it is clear that the net force down the inclined plane is equal to (Fg sin θ − fk), and so the acceleration is a=Fg sinθ−fk =(58.8N)sin39◦−5.4835N=5.25m/s2 m 6.0 kg Therefore, the box accelerates down the slope with a = 5.3 m/s2.

Josh punches his open left hand with his right hand. Which statement is true about the forces his two hands exert on each other?

Josh's right hand and left hand exert equal and opposite forces on each other.

Michael leans on a shopping cart, causing it to start moving. Which statement is true about the forces Michael and the cart exert on each other?

Michael and the cart exert equal and opposite forces on each other.

Miguel leans against a wall. Which statement is true about the forces Miguel and the wall exert on each other?

Miguel and the wall exert equal and opposite forces on each other.

Jack and Luis are pushing on opposite ends of a shopping cart. At the moment, the cart is staying at rest. We know that the forces each person exerts on the cart are equal in magnitude and opposite in direction. What is the reason this must be true?

Newton's second law tells us these two forces must be equal and opposite.

b. What will the acceleration be if the object's mass is doubled?

Next, if we double the mass (assuming the force remains the same), we get F1 F 1 1 2 2 apartb=2m=2 m =2 a =2 7.50m/s =3.75m/s Therefore, if we double the mass, we will get half the acceleration, equal to 3.75 m/s2.

What will the acceleration be if the force and the object's mass are both doubled?

Now, if we double both the force and the mass, we get 2F F 2 apartc=2m= m =a=7.50m/s Therefore, if we double both the force and the mass, we will get the same acceleration, equal to 7.50 m/s2.

How much force does the astronaut exert on his chair while accelerating straight up at 11.0 m/s2?

Solution: Again, we need the normal force, but this time the system is accelerating upward, so the normal force must be greater than the force of gravity on the astronaut. The net force on the astronaut is then the normal force ⃗n acting upward minus the force of gravity F⃗g because it is acting downward, and this net force can be set equal to the mass of the astronaut times the 11.0 m/s2 acceleration. Thus n−Fg =ma n=ma+Fg =ma+mg=m(a+g)=(92.0kg) 11.0m/s2+9.80m/s2 =1913.6N By Newton's third law, this normal upward force of the chair on the astronaut is equal to the downward force exerted by the astronaut on his chair. Therefore, the desired force exerted by the astronaut on his chair while accelerating upward at 11 m/s2 is 1910 N.

A stunt man drives a car at a speed of 20m/s off a 21 m high cliff. The road leading to the cliff is inclined upward at an angle of 20◦? (a) How far from the base of the cliff does the car land?

The car launches with v0 = 20 m/s at an angle of 20◦. Thus, its initial velocity along the x-direction is v0x = v0 cos 20◦, and its initial velocity along the y-direction is v0y = v0 sin 20◦, upward. We have enough information to find the time of flight from the y-direction motion, and then use this along with the constant velocity in the x-direction to find R, the distance from the base of the cliff at which the car lands. As I discussed in class, you don't have to waste precious time calculating the time to reach the maximum height and adding to the time to fall from maximum height to the base of the cliff. That's because the second equation of motion from the kinematic chapter involves the displacement. But you do have to get all the signs correct, otherwise it won't work. The initial y-velocity is positive, and the 21 m displacement is negative because it is in the downward direction. To show you the complete step, I'm going to write the complete displacement so that you see where this step comes from. Let the maximum height to which the car rises above the launch point be H. Then, the car rises up vertically by H, falls down vertically by −H, and then falls down a farther −21 m. Keep in mind that H is the maximum height above the launch point to which the car rises, not the maximum height from the base of the cliff. So we have H−H−21m=(v0 sin20◦)∆t+1(−9.80m/s2)(∆t)2 2 This gives us a quadratic equation: or Using the quadratic formula, we get −21 = (20 sin 20◦)∆t − 4.9(∆t)2 4.9(∆t)2 − 6.8404 ∆t − 21 = 0 ∆t = −(−6.8404) ± (−6.8404)2 − 4(4.9)(−21) = 2.8827 s, −1.4867 s 2(4.9) The second solution is unphysical, so we drop it. The time taken for the motion in the y-direction is then 2.8827 s. As usual, I'm retaining more digits than significant in these intermediate stages so I don't run into rounding errors. On the next page, I will solve for the range. V0 sin 20° PHY 170 (Autumn 2017) Homework 3 solutions: Page 3 On the previous page, we calculated that the time taken for the motion in the y-direction is 2.8827 s. Recall one of the two key points about projectile motion is that the ∆t for the y-direction is the same along the x-direction. In other words, you can treat the two motions, horizontal and vertical, independently of each other, but sharing a common time. Thus, along the x-direction, we have motion for ∆t = 2.8827 s with the constant velocity of v0x = (20 m/s) cos 20◦. The displacement along the x-direction is thus ∆x = (20 cos 20◦) ∆t + 1 (0)(∆t)2 2 so that ∆x = (20 m/s) cos 20◦ (2.8827 s) = 54.18 m Therefore, the distance from the base of the cliff at which the car lands is 54 m.

Blocks A and B have equal masses and are connected by a rope that passes over a frictionless pulley, as shown. At this moment, the system is in motion, with block B moving downward and A moving upward at a constant speed. Compare the magnitudes of the force of tension exerted on block A and the force of tension exerted on block B.

The force of tension exerted on block A is equal in magnitude to the force of tension exerted on block B.

Stephanie pulls on a rope that is attached to a large crate. She is pulling the crate along at a constant speed as it slides across rough carpet. How does the magnitude of the force of tension exerted on Stephanie compare to the magnitude of the force of tension exerted on the crate?

The force of tension exerted on the crate has the same magnitude as the force of tension exerted on Stephanie.

In the figure, a hand is pushing on block B to the right, causing both blocks to move to the right across a rough surface at a constant speed. Compare the force that pushes block B to the right to the force that pushes block A to the right.

The force that pushes block A to the right is smaller than the force that pushes block B to the right.

A 1300 kg steel beam is supported by two ropes.(a) What is the tension in rope 1?

The free-body diagram at the point where the steel beam is joined to the ropes is shown in the figure on the right. The three forces are shown in black: they are the force of gravity F⃗g on the steel beam, and the tensions T⃗1 and T⃗2 in the two ropes. The usual procedure is to decompose the appropriate forces, in this case T⃗1 and T⃗2, into their x and y components. I've marked the components in the figure in red: they are T1x and T1y, and T2x and T2y respectively. I've tried to draw the figure to stay approximately true to the sizes of the components — since the beam is in equilibrium, T1x and T2x must balance each other, so I've drawn their lengths to be about the same. Likewise, Fg must be equal to the sum of T1y and T2y, so I've tried to keep the vector representing F⃗g long enough for this to look visually plausible. The components of the tension forces can be found either by using the angles given, or using the angles made by these vectors with the x-axis but either way we have to get the same value: T1x = T1 sin 20◦(−ˆi) or T1x = T1 cos 70◦(−ˆi) I'll use the given angles, so the other components are T2x = T2 sin 30◦(ˆi), T1y = T1 cos 20◦(ˆj), T2y = T2 cos 30◦(ˆj) Since the beam is in equilibrium, the net force in the horizontal direction must be zero: ΣFx = 0. I've written the unit vectors in the equations for the components, so it's easy to keep track: ΣFx ≡ T2x − T1x = 0 or, equivalently T1x = T2x from which we get T1 sin 20◦ = T2 sin 30◦ (3.4.1) from which we get Equations (3.4.1) and (3.4.2) are two equations in two unknowns, so we can solve them to find T1 and T2. Likewise, the net force in the vertical direction must be zero: ΣFy = 0, and so ΣFy ≡T1y +T2y −Fg =0 or,equivalently T1y +T2y =Fg The problem is continued on the next page . . . T1 cos 20◦ + T2 cos 30◦ = Fg (3.4.2) T1 T1x T1y T2y 20° 30° T2x Fg T 2 Solution: We can solve for T1 in equation (3.4.2) if we eliminate T2 by using equation (3.4.1). From equation (3.4.1), we get T1 sin 20◦ T2 = sin 30◦ T1 cos20 + sin30◦ 0.939693 T1 + 0.592396 T1 = (1300 kg) 9.8 m/s2 cos30 =Fg (3.4.3) Substituting in equation (3.4.2), we get ◦ T1 sin20◦ Putting in the numbers so that ◦ T1 = (1300) 9.8 = 8315 N 1.532089 Therefore, the tension in rope 1 is 8300 N or 8.3 × 10>3 N.

Blocks A and B are connected by rope 1 and are pulled across a rough surface by rope 2, which is attached to block B, as shown. The applied force is strong enough to make the blocks speed up. Compare the magnitude of the tensions in ropes 1 and 2.

The magnitude of the tensions in the two ropes will not be equal.

You are standing in an elevator as it starts to move downward, accelerating for a moment. Which of these is true at that moment?

The net force on you changes from zero to downward as the elevator begins accelerating.

Consider a car driving at a constant speed on a level road. Which of the following is an action-reaction pair (from Newton's third law)?

The normal force exerted on the car by the road and the downward force exerted on the road by the car.

Consider a car speeding up as it drives along a level road. Which of the following is an action-reaction pair (from Newton's third law)?

The normal force exerted on the car by the road and the downward force exerted on the road by the car.

Allison pulls her little brother, Brody, in a sled on a snow-covered path at a constant speed. Which of the following is an action-reaction pair (from Newton's third law)?

The pulling force that Allison exerts on the sled and the pulling force that the sled exerts on Allison.

A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 120 m above the glacier at a speed of 150 m/s. At what horizontal distance from the target should the package leave the airplane?

The situation is sketched in the figure above. The initial velocity v0 = 150 m/s is entirely in the x-direction, assuming the plane is in level flight when it drops the package. Therefore, the initial velocity v0y in the y-direction is zero. R in the figure above is the horizontal distance from the target from where the package needs to leave the airplane. Let us consider the y-direction motion first. Starting from rest in the y-direction, the package of food falls through ∆y = −120 m in a time ∆t, which we can determine by using the second equation of motion under constant acceleration. Assuming free fall in the y-direction so that ay = −9.80 m/s2, as is the usual case in any projectile problem, the second equation of motion gives from which we obtain that −120 m = 0 + 1 (−9.80 m/s2) (∆t)2 2 2 (120 m) ∆t = 9.80 m/s2 = 4.9487 s Recall that ∆t is the same for the horizontal and vertical motion. In other words, if ∆t is the time taken by the package to fall from its drop point to the glacier surface, it is also the time to travel the distance R along the horizontal direction. But motion along the horizontal direction is at the constant velocity, v0 = 150 m/s; i.e., a = 0 along the x-direction. So, knowing ∆t, it is easy to find the displacement R using the second equation of motion: or R = v0∆t + 1(0)(∆t)2 2 R = v0∆t = (150 m/s) (4.9487 s) = 742.305 m Therefore, the horizontal distance from the target at which the package should leave the airplane is 742 m.

You are about to check out at the grocery store and you put a box of cereal on the conveyer belt at the register. The cashier turns on the conveyor belt, causing the box of cereal to speed up momentarily. At this moment, what can we say about the frictional force acting on the box of cereal?

There is a static frictional force pointing in the box's direction of motion, parallel to the conveyer belt.

What is the tension in rope 2?

This can now be found easily from equation (3.4.3) above: T1 sin 20◦ (8315 N) sin 20◦ T2 = sin 30◦ = sin 30◦ = 5688 N Therefore, the tension in rope 2 is 5700 N or 5.7 × 103 N.

How much force does an 92.0 kg astronaut exert on his chair while sitting at rest on the launch pad?

This might seem deceptively simple, but it is an important first step into the world of Newton's Third Law. The force exerted by the astronaut on his chair is an action-reaction pair to the normal force exerted perpendicular to the chair and upward on the astronaut by the chair. So, if we can find the normal force on the astronaut, we will be able to find the force exerted by the astronaut on his chair. Since the astronaut is sitting at rest on the launch pad, we can use Newton's first law to solve for the magnitude of this normal force. The only force opposing the normal is the force of gravity on the astronaut acting in the downward direction. Therefore, we have n = Fg, and so n = mg = (92.0 kg)(9.80 m/s2) = 901.6 N By Newton's third law, the force exerted by the astronaut on the chair while sitting at rest on the launch pad is equal to the normal force exerted by the chair on the astronaut. Therefore, the force exerted by the astronaut on the chair while sitting at rest on the launch pad is 902 N.

A 1500 kg car skids to a halt on a wet road where μk = 0.54. How fast was the car traveling if it leaves 62 m long skid marks?

This problem combines material from the chapter on force with material from the chapter on kinematics. Planning: We are given that the final velocity vf = 0, displacement ∆x = 62 m, and asked to find the initial velocity vi. We know from the equations of motion under constant acceleration that solving this problem would require us to know the acceleration a. So, is there a way to find a? We know the mass of the car, so we can find the force of gravity (Fg) on the car. In the vertical direction, the only force opposing Fg is the normal force n, as is clear from the free-body diagram below. And, we know that we can find the force of kinetic friction from the normal: fk = μkn, where μk is the coefficient of kinetic friction, whose value we are given. The force of kinetic friction is the only force along the horizontal direction, so if we divide it by the mass, we will have the acceleration of the car and we can then use the kinematic equations to solve for its initial velocity. So, onward! The force of gravity on the car is Fg = mg, where m = 1500 kg. Along the vertical direction, the only force opposing Fg is the upward-directed normal force. These two forces must balance since the system is in equilibrium along the vertical direction. Thus n = Fg We know that the force of kinetic friction is related to the normal force via fk =μkn where μk is the coefficient of kinetic friction; μk = 0.54 in this problem. Putting all this together, we have fk = μkn = μk(Fg) = μk(mg) = 0.54(1500 kg)(9.80 m/s2) = 7938 N and from Newton's second law a= F = 7938N =5.292m/s2 m 1500 kg We have our acceleration, ⃗a = −5.292 m/s2; I've inserted a minus sign because the acceleration must be in the opposite direction to the velocity to slow the car down to a stop. With vf = 0 and ∆x = 62 m inserted into the third equation of motion, we get so that (0)2 = vi2 + 2(−5.292 m/s2) 62 m Therefore, the car was traveling with a velocity of 26 m/s.

A constant force is applied to an object, causing the object to accelerate at 7.50 m/s2.

What will the acceleration be if the force is doubled? Solution: All parts of this problem are just a simple application of Newton's Second Law: ΣF = ma. As I discussed in class, I like to put the Σ before the F to remind myself that we're talking about the net force here, but since we're dealing with only one constant force, let's call it F in this problem. Now, to the problem. Let's isolate the acceleration to one side: a = F . m If we double the force (assuming the mass remains the same), we get 2F F 2 2 aparta = m =2 m =2 a =2 7.50m/s =15.0m/s Therefore, if we double the force, we will get twice the acceleration, equal to 15.0 m/s2.

Floating in deep space, you find yourself at rest next to a large asteroid. You reach out and tap the asteroid with a hammer. What happens to you in this process?

You will briefly accelerate away from the asteroid and then drift away at a constant speed.

The two blocks in the figure are sliding down the incline. What is the tension in the massless string?

You'll need two free-body diagrams to solve this problem, and I've drawn them above to the left of the physical picture of the system. The free-body diagram on the extreme left is for the 1.0 kg block, so I've included subscripts "1" in all the force labels. The free-body diagram in the middle to the immediate left of the physical picture of the two blocks is for the 2.0 kg mass, so I've put subscripts "2" in all the force labels. I've chosen the x-axis parallel to the inclined plane, as is usual in inclined-plane problems, and marked the x and y axes in each free body diagram as dashed blue lines. In both free-body diagrams, the normal force n acts perpendicular to the inclined plane, i.e., pointing toward the positive side of our chosen y-direction. Opposing this is the cosine component of Fg, which is acting toward the negative of our chosen y-direction. Therefore, we can write n1 = Fg1 cos 20◦ and n2 = Fg2 cos 20◦ (H4.3.1) Parallel to the inclined plane (i.e., along the chosen x-axis), we have the sine component of Fg acting down the inclined plane, and kinetic friction acting up the plane in both the free-body diagrams for the 1 kg and 2 kg masses respectively. Finally, the tension T of the string acts down the inclined plane for the 1 kg mass, and up the plane for the 2 kg mass. Before we can find T in the string, we must figure out what the system is doing. Is it accelerating or traveling at constant velocity? For the 1 kg mass, we have Fg1 sin 20◦ = (1.0 kg)(9.8 m/s2) sin 20◦ = 3.3518 N (H4.3.2) retaining more digits than significant for now, and fk1 = μ1n1 = 0.20 (1.0 kg)(9.8 m/s2) cos 20◦ = 1.8418 N (H4.3.3) where I've used equation (H4.3.1) to write the expression for n1. We haven't figured out the value of T yet, but we see that Fg1 sin 20◦ > fk1, and T makes that difference even greater because it is on the same side as Fg1 sin 20◦, so the 1 kg block must be accelerating as it moves down the incline. Assuming that the string remains taut and does not sag, the 2.0 kg block must also be accelerating down the incline with the same acceleration. x Fg1 cos 20° fk1 y n1 T Fg1 sin 20° 20° 20° Fg1 x fk2 T 20° y n2 Fg2 sin 20° 20° Fg2 cos 20° Fg2 On the next page, we'll find the value of T using this information. PHY 170 (Autumn 2017) Homework 4 solutions: Page 4 Problem 3 — continued On the previous page, we found that Fg1 sin 20◦ > fk1, meaning that the 1 kg mass, and by extension the whole system (assuming the string does not sag) is accelerating down the incline with an acceleration of, say, a. Since the tension T on the 1 kg mass acts down the incline, the sum of Fg1 sin 20◦ plus T minus fk1 must be equal to ma, by Newton's second law. That is Fg1 sin 20◦ + T − fk1 = m1a Using equations (H4.3.1)-(H4.3.3) on the previous page, we then get 3.3518 + T − 1.8418 = (1.0 kg) a 1.51 + T = a (H4.3.4) Equation (H4.3.4) has two unknowns, T and a, so we need another equation. We can get the other equation by considering the free-body diagram for the 2 kg mass. Applying the same considerations as above for the 2 kg mass, but with T acting up the incline this time, we get from Newton's second law that Fg2 sin 20◦ − fk2 − T = m2a where fk2 = μ2n2. Thus, writing n2 from equation (H4.3.1), we get (2.0 kg) (9.8 m/s2) sin 20◦ − 0.1(2.0 kg) (9.8 m/s2) cos 20◦ − T = (2.0 kg) a so that 6.7036 − 1.8418 − T = 2a 4.8618 − T = 2a or We can eliminate a from equations (H4.3.4) and (H4.3.5) to find the value of T . Put the value of a from equation (H4.3.4) into equation (H4.3.5): 4.8618 − T = 2 1.51 + T which becomes and may be written as so that and, therefore 4.8618−T = 3.02+2T 3T = 4.8618 − 3.02 3T = 1.8418 T = 1.8418 = 0.613 N 3 (H4.3.5) Therefore the tension in the massless string is 0.61 N.

In general, when a car is driving along a straight road, the force of friction between a car tire and the road _____________.

can point either forward or backward, depending on what the car is doing.

The mass of an object

is not a vector and has no direction.

b. What is the car's impact speed?

ou have to be careful here, because many students forget about the horizontal velocity! The car travels at constant velocity in the horizontal direction, so its impact velocity along the x-direction is the same as that at launch, namely v0 cos 20◦. Meanwhile, we can use the first equation of motion from kinematics to calculate its velocity in the y-direction: so that (vy)at impact = v0y + (−9.80 m/s2) (∆t) (vy)atimpact =(20sin20◦)+(−9.80m/s2)(2.8827s)=−21.410m/s The answer comes out with a negative sign, which makes sense because the y-direction velocity is downward at impact. The impact speed is then so that v = (vx)2at impact + (vy)2at impact v = (20 cos 20)2 + (−21.410)2 = 28.4 m/s Therefore, the car's impact speed is 28 m/s.

A person is trying to pull a heavy crate across a room with a rope that is attached to the crate, but it doesn't move. The tension in the rope

s equal at both ends of the rope

A box rests on a table. The normal force exerted by the table on the box is equal in magnitude to the weight of the box and points in the opposite direction. This is because

the net force on the box is zero and, therefore, the forces must be balanced

Consider a car at rest, parked on level ground. The force of friction between a car tire and the ground

zero