What is the Work Done on a Block by Different Forces?

[Jacob87411]

Having some problems with this question..any help is appreciated

A 14.0 kg block is dragged over a rough, horizontal surface by a 86.0 N force acting at 20.0° above the horizontal. The block is displaced 5.00 m, and the coefficient of kinetic friction is 0.300.

(a) Find the work done on the block by the 86.0 N force.
W=Fd so work=(86.0 N) ( 5m) =
430

(b) Find the work done on the block by the normal force.
Again this will be the force * distance. The force here is the normal force which is given by mgsin(20) then multiplied by the distance of 5?

(c) Find the work done on the block by the gravitational force.
Is this just the weight times the distance traveled...so sin(20)*mg*5

 

[Hootenanny]

(a) You need to resolve the force in the horizontal plane first. Work done is the force multiplied by the distance moved in the direction of that force

 

[Hootenanny]

(b) and (c) You need to think about all the forces acting in the vertical plane. As there in no movement in the vertical plane, the sum of the forces acting upwards must equal mg. However, this is irrelevent, work is only done when a force is moved through a distance is the direction of that force.

 

[VietDao29]

Having some problems with this question..any help is appreciated

A 14.0 kg block is dragged over a rough, horizontal surface by a 86.0 N force acting at 20.0° above the horizontal. The block is displaced 5.00 m, and the coefficient of kinetic friction is 0.300.

(a) Find the work done on the block by the 86.0 N force.
W=Fd so work=(86.0 N) ( 5m) =
430

(b) Find the work done on the block by the normal force.
Again this will be the force * distance. The force here is the normal force which is given by mgsin(20) then multiplied by the distance of 5?

(c) Find the work done on the block by the gravitational force.
Is this just the weight times the distance traveled...so sin(20)*mg*5

Nah, you should look up the equation for work again, it's:
$$W = \vec{F} . \vec{s} = |F| |s| \cos \varphi$$
where $$\varphi$$ is the angle between the force and the displacement vector.
In (a) $$\varphi = 20 ^ \circ$$. What's $$\varphi$$ in (b), and (c)?