Work done on a block by net force

AI Thread Summary
The discussion focuses on calculating the work done on a 16.9 kg steamer trunk being dragged with a force of 177 N at an angle of 34° over a distance of 59.3 m, considering a coefficient of kinetic friction of 0.186. Participants emphasize the importance of drawing a free-body diagram to identify the forces acting on the trunk, including weight, normal force, friction, and the applied force. The net force (Fnet) is derived from the balance of vertical forces and the frictional force, which is calculated using the coefficient of kinetic friction. Clarifications are made regarding the relationship between the reaction force and the normal force, confirming they are equivalent in this context. The conversation aims to guide the calculation of work done against friction as the trunk is moved.
kt7477
Messages
2
Reaction score
0
1. You drag a(n) 16.9 kg steamer trunk over a
rough surface by a constant force of 177 N
acting at an angle of 34◦ above the horizontal.
You move the trunk over a distance of 59.3 m
in a straight line, and the coefficient of kinetic
friction is 0.186.
The acceleration of gravity is 9.8 m/s2 .How much is the work done on the block by
the net force?
Answer in units of J.

u= coefficiant of finetic friction
@= theta
Fnet= [sqrt (mgsin@^(2)+177^(2))] -uFncos@

W= Fnet (d)
W= ? * (59.3)

I have trouble figuring out how to calculate Fnet.
 
Physics news on Phys.org
I would start by drawing a diagram of the situation. Draw the 4 forces in, i.e. weight W, reaction force R from the floor, friction F' and the force we are pulling with F.

Since the trunk isn't moving the vertical direction, we may equate the 'downwards' forces with the 'upwards' forces, i.e.

W=R+Fsin(@)

Also we know F'=uR, where u is the coeff of kinetic friction, so we may rearrange the above to find the frictional force in terms of the F, W and u.

It is this frictional force that we do work against, so the work is F' times s, the distance through which we move it trunk.
 
is reaction force the same as normal force?
 
Yes it is in this case. It is pointing vertically upwards.
 
Welcome to PF!

kt7477 said:
is reaction force the same as normal force?
Rudipoo said:
Yes it is in this case. It is pointing vertically upwards.

Hi kt7477! Welcome to PF! :smile:

Technically, the reaction force is the total force of the block on the ground, or vice versa.

So it's the normal force plus the friction force.

It's best just to call the normal force "the normal force". :wink:

(sorry if that seems unnecessarily complicated :smile:)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top