A 4.6 kg object is moving at 7.2 m/s. A 6.9 N force is applied in the direction

Question

A 4.6 kg object is moving at 7.2 m/s. A 6.9 N force is applied in the direction of motion and then removed after the object has travelled an additional 2.4 m. How much work was done by this force? A 3056 N car is initially travelling at 21 m/s. The brakes are applied and it comes to a stop in 1.1 s. How much kinetic energy did it lose in this time? A man pulls a 81 N crate up a frictionless 9 degree slope starting at the bottom and going all tire way to the top. The slope is 3.5 m high Assuming that the crate moves at a constant speed, how much work did the man do on the crate?

Explanation / Answer

4.solution

work is calculated by force times the distance moved in the same direction of the force.

this problem has a lot of extra information.

the answer is force x distance or 6.9N x 2.4m = 16.56J

5.solution

Given the equations

KE = 0.5 * m * v^2

W = m * g

Given the values

W = 3056 N
Vi = 21 m/s
Vf = 0 m/s
t = 1.1 s
g = 9.81 m/s^2

First find mass of the car

3056 N = m * (9.81 m/s^2)

Mass car = 311.836 kg

Find the original kinetic energy

KE = 0.5 * (311.836 kg) * (21 m/s)^2
KE = 68759.83 J

3.solution

h is the height. In this problem they clearly state that the height is 3.5m so

W= mgh = 81x3.5= 283.5J

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