How Does Force Direction Affect Work Done?

[Tdollarhyde]

A force of F = 35 N is used to drag a crate 7 m across a floor. The force is directed at an angle upward from the crate so that the vertical component of the force is Fv = 21 N and the horizontal component is Fh = 28 N as shown in the diagram.

(a) What is the work done by the horizontal component of the force?
196 Joules
(b) What is the work done by the vertical component of the force?
? Joules
(c) What is the total work done by the 35 N force?
? Joules

Can't figure out how to do the last 2 for the life of my. You don't even have to do it for me, I'll be good with just an equation.

Also, Here is another one I can't figure out:

An ice skater with a mass of 89 kg pushes off against a second skater with a mass of 30 kg. Both skaters are initially at rest.
(a) What is the total momentum of the system after they push off?
? kg·m/s
(b) If the larger skater moves off with a speed of 1.8 m/s, what is the corresponding speed of the smaller skater?
? m/s

 

[Tdollarhyde]

Wow, I'm stupid. The vert. force is doing 0 work, and the overall work is the same as the horizontal work!

 

[PhanthomJay]

Not stupid, just an oversight. Try the 2nd question and show an attempt if you still need help with that one.

 

[Tdollarhyde]

Not stupid, just an oversight. Try the 2nd question and show an attempt if you still need help with that one.

Well, the total momentum is 0, and the second skater was going somewhere around 5 something. Have since put it away.

 

[rwisz]

(a) The equation for work done is W = F * d, where W is work, F is force, and d is distance. Since we are given the force (F = 35 N) and distance (d = 7 m), we can simply plug these values into the equation to find the work done by the horizontal component of the force:

W = (28 N) * (7 m) = 196 Joules

(b) Similarly, we can use the same equation to find the work done by the vertical component of the force:

W = (21 N) * (7 m) = 147 Joules

(c) To find the total work done by the 35 N force, we can use the Pythagorean theorem to find the magnitude of the force:

F = √(Fh^2 + Fv^2) = √(28^2 + 21^2) = √(784 + 441) = √1225 = 35 N

Then, we can use the same equation as before to find the total work done:

W = (35 N) * (7 m) = 245 Joules

(a) The total momentum of the system after they push off can be found using the equation p = m*v, where p is momentum, m is mass, and v is velocity. Since both skaters were initially at rest, their initial momentum is 0. After they push off, their total momentum will be equal and opposite, so the total momentum of the system is also 0.

(b) We can use the conservation of momentum to find the velocity of the smaller skater. Since the total momentum of the system is 0, the momentum of the smaller skater must be equal in magnitude but opposite in direction to the momentum of the larger skater. So we can set up the equation:

(m1 * v1) + (m2 * v2) = 0

where m1 and v1 are the mass and velocity of the larger skater, and m2 and v2 are the mass and velocity of the smaller skater. We can plug in the values we know and solve for v2:

(89 kg * 1.8 m/s) + (30 kg * v2) = 0

160.2 kg·m/s + 30 kg·m/s = 0

190.2 kg·m/s =