physics....question?

Question

To pull a 50kg crate across a frictionless floor a worker applies a force of 210N, directed at 20 degrees above the horizontal. As the crate moves 3m, what work is done on the crate by...
(a) the workers force
(b) the weight of the crate
(c) the normal force exerted by the floor on the crate
(d) total work done on the crate

i have letter (d) is it w=(210N)(3) (cos20) right? or is that for (a)?
also, is the nornal force 9.81*50kg

Answer 1

Your answer is for (a). There are certain assumptions, in this problem. It is assumed that the crate is pulled through a distance of 3 m without increasing its velocity or almost at rest. So net work or total work done on the crate has to be zero. So whatever you have calculated isthe work done by the worker on the crate. Negative work is done on the crate by the frictional force tangential to the crate or acting in the opposite direction of its motion which the floor applies on the crate. The answers to b and c are zero because the motion of the crate is perpendicular to the force in both cases and cos 90 is zero.

Answer 2

a)210N
b)50kg
c)fx=f(cos@)
f= 210 (cos20)
f=197.34 N

d)work done = force x distance
work done =197.34 (3)
=592.02





I DONT KNOW IF IM CORRECT IM JUZ TRYING OKAY...