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

In summary: In (a) \varphi = 20 ^ \circ. What's \varphi in (b), and (c)?In (b), \varphi = 20 ^ \circ. and in (c), \varphi is the angle between the force and the displacement vector.
  • #1
Jacob87411
171
1
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
 
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  • #2
(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
 
  • #3
(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.
 
Last edited:
  • #4
Jacob87411 said:
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:
[tex]W = \vec{F} . \vec{s} = |F| |s| \cos \varphi[/tex]
where [tex]\varphi[/tex] is the angle between the force and the displacement vector.
In (a) [tex]\varphi = 20 ^ \circ[/tex]. What's [tex]\varphi[/tex] in (b), and (c)?
 

What is work done on a block?

Work done on a block refers to the amount of energy transferred to or from the block by forces acting on it. It is a measure of the change in the block's kinetic or potential energy.

How is work calculated?

Work is calculated by multiplying the force applied to the block by the distance it is moved in the direction of the force. This can be represented by the equation W = F*d, where W is work, F is force, and d is distance.

What is the unit of measurement for work?

The unit of measurement for work is joules (J). One joule is equal to one newton-meter, which means that one joule of work is done when a force of one newton is applied to an object and it is moved one meter in the direction of the force.

Can negative work be done on a block?

Yes, negative work can be done on a block. This occurs when the force applied to the block is in the opposite direction of its motion, resulting in a decrease in the block's kinetic or potential energy. For example, if a block is pushed up a hill and then rolls back down, the force of gravity does negative work on the block as it moves downhill.

How is work related to power?

Power is the rate at which work is done. It is calculated by dividing the amount of work by the time it takes to do that work. This can be represented by the equation P = W/t, where P is power, W is work, and t is time. In other words, power measures how quickly work is being done on a block by forces.

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