1. A woman wishes to pull a 3.9 kg crate 6 m across a rough floor. She applies a
ID: 2254813 • Letter: 1
Question
1. A woman wishes to pull a 3.9 kg crate 6 m across a rough floor. She applies a 102 N force, and she will pull so that the rope is parallel to the floor. The coefficient of kinetic friction between the crate and the floor is k = 0.5. Calculate the total work done on the crate as it moves the 6 m distance.
2. A 4.4 kg object is moving at 2.5 m/s. A 1.1 N force is applied in the direction of motion and then removed after the object has traveled an additional 3.6 m. How much work was done by this force?
3.A disk with a moment of inertia of 5.9 kg·m2 and a radius of 1.1 m rotates on a fixed axis perpendicular to the disk and through its center. A force of 5.4 N is applied tangentially to the rim. Calculate the work done by the torque as the disk turns through half a revolution.
4.A 3665 N car is initially traveling at 5.3 m/s. The brakes are applied, and it comes to a stop in 4.7 s. How much kinetic energy did it lose in this time?
5.A man pulls a 72 N crate up a frictionless 22.5o slope, starting at the bottom and going all the way to the top. The slope is 6.4 m high. Assuming that the crate moves at a constant speed, how much work did the man do on the crate?
6.A 8.0g bullet is fired horizontally into a 100kg block of wood that is suspended by a rope from the ceiling. The block swings in an arc, rising 55.8cm vertically above its lowest position. What was the speed of the bullet just before the collision?
7.Blocks A and B are moving toward each other along the x-axis. Block A has a mass of 2.0 kg and a velocity of +50 m/s; block B has a mass of 4.0 kg and a velocity of -25 m/s. They collide and undergo a perfectly elastic collision and move off along the x-axis. How much kinetic energy is transferred from A to B in this collision?
8.A boy holds a 40 N weight at arm's length, 0.7 m from his shoulder. His arm is 1.5 m above the ground. He holds the weight in this position for 2.5 s. How much work did he do on the weight during this time?
show work pleace
Explanation / Answer
1) work done= work done on carate + frictional work done = force*displacement + frictional force*displacement
=102*6 + 0.5*3.9*9.81*6 = 726.77J
2)work done=force*displacement = 1.1*3.6 = 3.96J
3)W = F ? r= 5.4*pie*1.1 = 19.34J
4)kinetic energy lost = 0.5*mass*(V-v)^2 = 0.5*3665/9.81(5.3-4.7)^2 = 67.24J
5)since it moves with a constant velocity work done = potential energy = mass*g*height = 72*6.4= 460.8J
6)conserving energy for this we have
initial energy =0.5*mass of bullet+block*velocity of bullet^2 = 0.5*100.008*V^2
final energy = mass*g*height = 100.008*9.81*0.558 = 547.4J
so equating both wehave 0.5*100.008*V^2 = 547.4
V=3.3m/s
7)we have conservation of momentum
initial KE = Final KE
2*50 + 4*(-25) = 0
so both blocks will come to rest and no KE is transfered.
8)work done = force*displacemet
since there is no displacemebt so theer is no work done
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