Chapter 10, Problem 57A

a.

To determine

Work done by M in sliding the crate.

Answer to Problem 57A

Work done by M in sliding the crate up the ramp is $W=800\text{J}$

Explanation of Solution

Given:

Force parallel to ramp is $F=400\text{N}$

Mass of crate is $m=60\text{kg}$

Length of ramp is $d_1=2\text{m}$

Height of ramp is $d_2=1\text{m}$

Formula used:

Since force is in the direction of motion i.e. along the ramp.

So, work done in sliding the crate up is given by,

  $W=F{d}_1$ …. (1)

Calculation:

On putting the given values in equation (1)

  $\begin{array}{l}W=400\times 2\\ =800\text{J}\end{array}$

b.

To determine

Work done by M in lifting the crate up.

Answer to Problem 57A

Work done by M in lifting the crate up is $W=588\text{J}$

Explanation of Solution

Given:

Force parallel to ramp is $F=400\text{N}$

Mass of crate is $m=60\text{kg}$

Length of ramp is $d_1=2\text{m}$

Height of ramp is $d_2=1\text{m}$

Formula used:

For lifting the crate only vertical displacement counts.

So, work done in lifting the crate up is given by,

  $W=F{d}_2$ …. (1)

Calculation:

As, $F=mg$

So, $W=mg{d}_2$

On putting the given values

  $\begin{array}{l}W=60\times 9.8\times 1\\ =588\text{J}\end{array}$