Question

A trunk of mass 22 kg is on the floor. The trunk has a very small initial speed. The acceleration of gravity is 9.8 m/s 2 .What constant horizontal force pushing the trunk is required to give it a velocity of 10 m/s in 20 s if the coefficient of sliding friction between the trunk and the floor is 0.51? Answer in units of N.

Answer 1

`F=120.96N`

Step-by-step explanation:

Given the information from the exercise we need to use Newton's Laws to solve it

If we analyze the y-axis

∑
`F_(y)=N-W=0`

`N=W=m*g`

Since the trunk is moving in the x-axis and its velocity is changing with time, its acceleration is:

`a=(10m/s)/(20s)=0.5m/s^2`

Knowing that we can analyze the forces acting in the horizontal direction

∑
`F_(x)=F-fr=m*a`

`F=m*a+fr`

`F=m*a+uN=m*a+u*m*g`

`F=(22kg)(0.5m/s^2)+(0.51)(22kg)(9.8m/s^2)=120.96N`

So, the force required to push the trunk is 120.96 N