Chapter 10, Problem 58A

a.

To determine

Work done by worker.

Answer to Problem 58A

Work done by worker is $340\text{J}$

Explanation of Solution

Given:

Horizontal force is $F=85\text{N}$

Displacement in the direction of force is $d=4\text{m}$

Formula used:

Work done is given by,

  $W=Fd$ ….. (1)

Calculation:

Putting the values in equation (1)

  $\begin{array}{l}W=85\times 4\\ =340\text{J}\end{array}$

b.

To determine

Work done by gravity.

Answer to Problem 58A

Work done by gravity is $-279\text{J}$

Explanation of Solution

Given:

Vertical displacement is 3.0m.

Weight of crate is $F=93\text{N}$

Formula used:

Work done by gravity is given as,

  $W=Fd$ ….. (2)

Calculation:

As displacement is opposite to gravity,

So, from (2)

  $\begin{array}{l}W=-Fd\\ =-\left(93\times 3\right)\\ =-279\text{J}\end{array}$

c.

To determine

Energy transformed by friction.

Answer to Problem 58A

Energy transformed by friction $-125.4\text{J}$

Explanation of Solution

Given:

Length of plane is 5.0m.

Weight of crate is $F=93\text{N}$

Horizontal force is $F=85\text{N}$

Coefficient of friction is $\mu =0.2$

Formula used:

Energy transformed by friction is,

  $W=\mu F_{N}d$ …. (3)

Calculation:

From (3)

  $\begin{array}{l}W=\mu \left(F_{\text{worker }\perp }+F_g{}_{\perp }\right)d\\ =0.2\left(85\sin \theta +93\cos \theta \right)\left(-5\right)\left(\because \text{displacment is opposite to friction}\right)\\ =0.2\left(85\times \frac{3}{5}+93\times \frac{4}{5}\right)\left(-5\right)\\ =-125.4\text{J}\end{array}$