PHY101 - Mastering Physics #4
A person places a cup of coffee on the roof of her car while she dashes back into the house for a forgotten item. When she returns to the car, she hops in and takes off with the coffee cup still on the roof. If the coefficient of static friction between the coffee cup and the roof of the car is 0.27, what is the maximum acceleration the car can have without causing the cup to slide? Ignore the effects of air resistance. What is the smallest amount of time in which the person can accelerate the car from rest to 24 m/s and still keep the coffee cup on the roof?
A) In the vertical: Fnety = 0 N - mg = 0 N = mg In the horizontal: Fnet = ma Friction = ma Uk(N) = ma Uk(mg) = ma Ukg = a .27*9.8 = a a =2.646 m/s^2 *the minimum acceleration to keep it up top B) a = (vf - vi) / time 2.646 = (24 - 0) / time t = 9.07s
On vacation, your 1350-kg car pulls a 600-kg trailer away from a stoplight with an acceleration of 1.80 m/s^2. What is the net force exerted by the car on the trailer? Assume that the positive x axis is directed toward the direction of motion. What force does the trailer exert on the car? Assume that the positive x axis is directed toward the direction of motion. What is the net force acting on the car? Assume that the positive x axis is directed toward the direction of motion.
A) F = (600 kg) * (1.80 m/s^2) = 1080 N B) -1080 N C) Ftotal=(1350+600)*1.80 Ftotal = 3510 N Ftotal - F = 3510 N - 1080 N = 2430 N
When you lift a bowling ball with a force of 90 N , the ball accelerates upward with an acceleration a. If you lift with a force of 98 N , the ball's acceleration is 2a. Find the weight of the bowling ball. Express your answer using two significant figures. Find the acceleration a. Express your answer using two significant figures.
A) F1 - mg = ma F2 - mg = m(2a) ma = F1 - mg = 1/2 * (F2-mg) 2*F1 - F2 = mg m = [2(90) - (98)]/9.8 = 8.4 kg w = 8.4 kg * 9.81 m/s^2 = 82 N B) a = [F1 - mg]/m = [90N - (8.4 kg * 9.81 m/s^2)]/(8.4 kg) =0.96 m/s^2
An ant walks slowly away from the top of a bowling ball, as shown in the figure. If the ant starts to slip when the normal force on its feet drops below one-half its weight, at what angle θ does slipping begin? Express your answer as an integer.
Ant's Weight: W = mg Normal force at slip: N = mg(cosθ) According to the problem, normal force at slip = 1/2 weight N = (1/2)W mg(cosθ) = (1/2)mg or: cosθ = 1/2 cos-1 = 60
A heavy crate is attached to the wall by a light rope, as shown in the figure. Another rope hangs off the opposite edge of the box. If you slowly increase the force on the free rope by pulling on it in a horizontal direction, which rope will break? Ignore friction and the mass of the ropes.
Both ropes are equally likely to break. Since the attached rope doesn't have to support any weight, the tension is the same in both ropes.
What is the minimum horizontal force F needed to make the box start moving in (Figure 1) ? The coefficients of kinetic and static friction between the box and the floor are 0.28 and 0.35, respectively. 31 N going down 25 kg box
Find normal force (sum of all the vertical components of force): (25 kg * 9.81 m/s^2) + 31 N= 276 N Find frictional force: Ff = 276 N * (0.35) Ff = 97 N
A child goes down a playground slide with an acceleration of 1.34 m/s^2. Find the coefficient of kinetic friction between the child and the slide if the slide is inclined at an angle of 29.0 below the horizontal.
m * a = m*g*sinθ - u*m*g*cosθ 1.34 = g*sin(29) - u*g*cos(29) u = 0.398
A major-league catcher gloves a 92 mi/h pitch and brings it to rest in 0.15 m. If the force exerted by the catcher is 803 N, what is the mass of the ball? Express your answer using two significant figures.
v^2 = v0^2 + 2a(x-x0) a = (v2 − v02)/ [2 (x-x0)] v0 = 92 mi/h = 41.14 m/s a = (0 - (41.12)^2)/[2 * (0.15 - 0)] = -5641 m/s^2 F = ma a = |F/a |= |(803 N) / (-5641 m/s2)| = 0.14kg
To move a large crate across a rough floor, you push on it with a force F at an angle of 21 ∘ below the horizontal, as shown in the figure (Figure 1). Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.55. Express your answer using two significant figures.
ΣFx = F*cosθ - u*N = 0 ΣFy = N - mg - F*sinθ = 0 N = mg + F*sinθ F*cosθ - u(mg + F*sinθ) = 0 F*cosθ - u*F*sinθ = umg F = (umg)/(cosθ - u*sinθ) F = (0.55*32*9.81)/(cos21 - (0.55*sin21)) F = 237 N
A shopper pushes a 7.5-kg shopping cart up a 13 degree incline, as shown in the figure (Figure 1) . Find the magnitude of the horizontal force, F , needed to give the cart an acceleration of 1.55 m/s^2 . Express your answer using two significant figures.
∑Fx = F*cos θ - mg sin θ = ma F = m(a + g*sin θ)/(cos θ) F = (7.5 kg)*[(1.55 m/s^2) + (9.81 m/s^2)*sin(13)]/(cos(13)) F = 29 N
