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

AI Thread Summary
The discussion revolves around calculating the work done on a 14.0 kg block being dragged by an 86.0 N force at a 20° angle over a distance of 5.00 m, with a coefficient of kinetic friction of 0.300. For part (a), the work done by the applied force is calculated using the formula W = Fd, resulting in 430 J. In parts (b) and (c), participants emphasize the importance of considering the angle between the forces and the direction of displacement, noting that work is only done when a force moves through a distance in its direction. The normal and gravitational forces require further analysis to determine their contributions to work, as they act vertically while the displacement is horizontal. Understanding the relationship between force, distance, and angle is crucial for accurately calculating work done in physics.
Jacob87411
Messages
170
Reaction score
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
 
Physics news on Phys.org
(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
 
(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:
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:
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)?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Thread 'Struggling to understand the concept of the question (ice cube in water optics question)'

TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
View full post »

Similar threads

Finding the work done by a block
Replies
4
Views
3K
Minimum work done to push the cube over
Replies
14
Views
992
Force pulling on a string at an angle with a block at the other end
Replies
2
Views
1K
How do I find the work done for a 100 N downward force?
Replies
4
Views
2K
How can "work" done by a force be negative?
Replies
19
Views
1K
My Mistake? Understanding Friction Force & Work Done on Snowy & Icy Surfaces
Replies
29
Views
2K
Work energy theorem problem -- Block sliding up a curved incline
Replies
10
Views
1K
Calculating Work and Kinetic Energy on an Inclined Plane
Replies
2
Views
3K
Total work done while pushing a wheelchair
Replies
4
Views
2K
Work done during a collision -- Change in Kinetic Energy & change in Momentum
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top