# What is the velocity of the box after 3.1 s, the distance moved by the box and the average force exerted by the man in the following case:A man pushes a 24 kg box a cross a frictionless floor. The...

What is the velocity of the box after 3.1 s, the distance moved by the box and the average force exerted by the man in the following case:

A man pushes a 24 kg box a cross a frictionless floor. The box is initially at rest. He initially pushes on the box gently, but gradually increases his force so that the force he exerts on the box varies in time as F = 8.1*t N. After 3.1s, he stops pushing the box. The force is always exerted in the same direction.

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### 1 Answer

The man pushes a 24 kg box across a frictionless floor with a variable force that is equal to F = 8.1*t at any moment of time t.

The box is pushed for 3.1 s and is initially at rest. The acceleration of the box due to force applied is equal to 8.1*t/24

If the box has a velocity v at the start of an infinitesimally small time period dt, at the end of the time period the velocity is v + (8.1)*t/24*dt

To estimate the velocity of the box after 3.1 s, compute the definite integral of 8.1*t/24 between t = 0 and t = 3.1

V = `int_(0)^3.1 8.1*t/24 dt`

=> `(8.1/24)*(t^2/2)` between t = 0 and t = 3.1

=> `(8.1/48)(3.1^2 - 0)`

=> 1.621 m/s

The distance traveled in 3.1 s is the integral

`int_(0)^3.1 8.1/24)*(t^2/2) dt`

=> `int_(0)^3.1 (8.1/48)*(t^3/3) dt`

=> `(8.1/144)*3.1^3`

=> 1.675 m

The average force exerted by the man is (8.1*3.1 - 0)/3.1 = 8.1 N