Question

To push a 25.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel to the incline. As the crate slides 1.50 m, how much work is done on the crate by:_________
(a) the worker’s applied force,
(b) the gravitational force on the crate, and
(c) the normal force exerted by the incline on the crate?
(d) What is the total work done on the crate?

Answer 1

Answer:

a.  $W_w=313.5\ J$

b. $W_g=-155.312\ J$

c. $F_N=222.045\ N$

d.  $W_t=313.5\ J$

Explanation:

Given:

• angle of inclination of the surface,$\theta=25^{\circ}$

• mass of the crate, $m=25\ kg$

• Force applied along the surface, $F=209\ N$

• distance the crate slides after the application of force, $s=1.5\ m$

a.

Work done by the worker who applied the force:

$W_w=F.s\ cos 0^{\circ}$ since the direction of force and the displacement are the same.

$W_w=209\times 1.5$

$W_w=313.5\ J$

b.

Work done by the gravitational force:

$W_g=m.g\times h$

where:

g = acceleration due to gravity

h = the vertically downward displacement

Now, we find the height:

$h=s\times sin\ \theta$

$h=1.5\times sin\ 25^{\circ}$

$h=0.634\ m$

So, the work done by the gravity:

$W_g=25\times 9.8\times (-0.634)$  ∵direction of force and displacement are opposite.

$W_g=-155.312\ J$

c.

The normal reaction force on the crate by the inclined surface:

$F_N=m.g.cos\ \theta$

$F_N=25\times 9.8\times cos\ 25$

$F_N=222.045\ N$

d.

Total work done on crate is with respect to the worker: $W_t=313.5\ J$