Question

A factory worker pushes a 30.0 kg crate a distance of 4.3 m along a level floor at constant velocity by pushing downward at an angle of 30∘ below the horizontal. The coefficient of kinetic friction between the crate and floor is 0.25.

Answer 1

Answer:

a) the worker must apply 73.5 Newtons

b) work done = 330.75 Joules

c) work done = -330.75 Joules

d)

Work done by normal force = 0 Joules

Work done by gravity = 0 Joules

e) total work done = 0 joules

Step-by-step explanation:

The complete question:

"A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25. (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?"

a)

distance, s = 4.3

mass, m = 30

Coefficient of friction, $\mu_k=0.25$

The worker here must apply a force, $F_{worker}$ equal to kinetic frictional force $F_k$.

We know  $F_k=\mu_k mg$

We know g = 9.8, thus plugging in we get:

$F_{worker}=F_k=\mu_k mg=0.25*30*9.8=73.5$

Thus, the worker must apply 73.5 Newtons

b)

Work is given by the formula:

$W=F_{worker} s Cos\phi$

F and s are in the same direction, so angle ($\phi$)  is 0

Now we find work done:

$W=F_{worker} s Cos\phi\\W=73.5*4.5*Cos(0)\\W=73.5*4.5\\W=330.75$

Hence, work done = 330.75 Joules

c)

We use same formula but remember that the angle is 180 degrees since frictional force is acting against it (opposite).

Thus we have:

$W=F_{worker} s Cos\phi\\W=73.5*4.5* Cos(180)\\W=-330.75$

Hence, work done = -330.75 Joules

d)

Work on by normal force would be perpendicular, so angle would be 90 degrees. THus

$W_n=F_{worker} s Cos (90)\\W_n=0$

Work by gravity would be opposite of that direction. That would be -90 degrees. Thus

$W_g=F_{worker} s Cos(-90)\\W_g=0$

Thus

Work done by normal force = 0 Joules

Work done by gravity = 0 Joules

e)

Net work done on crate will be sum of all individual work done.

$W_{net}=W{worker}+W_k+W_n+W_g$

Plugging in the values, we have:

$W_{net}=W{worker}+W_k+W_n+W_g\\W_{net}=330.75-330.75+0+0=0$

Hence

total work done = 0 joules

Note: this is 0 because F_worker and F_k are working opposite to each other