help please

[badboyben03]

i need help with figureing out this problem

A sled of mass 57 kg is pulled along snow-covered, flat ground. The static friction coefficient is 0.30, and the sliding friction coefficient is 0.10.

What force will be needed to start the sled moving?

Once moving, what total force must be applied to the sled to accelerate it 4.6 m/s2?

What force is needed to keep the sled moving at a constant velocity?

I figured out the weight, but i need the formula to figure out the rest.

[deltabourne]

Originally posted by badboyben03
i need help with figureing out this problem

A sled of mass 57 kg is pulled along snow-covered, flat ground. The static friction coefficient is 0.30, and the sliding friction coefficient is 0.10.

What force will be needed to start the sled moving?

Once moving, what total force must be applied to the sled to accelerate it 4.6 m/s2?

What force is needed to keep the sled moving at a constant velocity?

I figured out the weight, but i need the formula to figure out the rest.
To start the sled moving, you need to overcome static friction, the force of which is mu*F_n. F_n in this case is equal to the mass*gravity of the sled.

So the force to start = mu*m*g = .3*57*9.81 = 167.751 N

Now to accelerate it to 4.6 m/s^2, you need to opposte the force of sliding (kinetic) friction

Net F_x = F_applied - F_friction = ma
F_applied is what we're looking for, F_friction again is mu*m*g where mu is KINETIC friction this time (not static)

So F_applied = mu_k*m*g+m*a (a is 4.6 m/s^2, the acceleration we want to get it to)
F_applied = .1*57*9.81+57*4.6 = 318.117 N

[badboyben03]

thanks. do you think you can help me w/ my last problem?

[deltabourne]

edit: whoops, didn't see it

To keep something at constant velocity the net Force must be zero.. so here we have F_applied - F_friction = 0

or F_applied = F_friction = mu_k*m*g = .1*57*9.81 = 55.917

[badboyben03]

A force of 45 N accelerates a 6.0-kg block at 7.0 m/s2 along a horizontal surface.

How large is the frictional force and what is the coefficient of friction?

[deltabourne]

Originally posted by badboyben03
A force of 45 N accelerates a 6.0-kg block at 7.0 m/s2 along a horizontal surface.

How large is the frictional force and what is the coefficient of friction?
ok.. again, just set up your net f_x forces. We have the applied force (let's call it F_a) and the frictional force (F_f)

Net F_x = F_a - F_f = m*a
Solving for F_f, we have F_a - m*a = F_f

The applied force is 45N, the mass is 6.0kg, and acceleration is 7.0 m/s^2.. plug them in, you get a frictional force of 3N

Now, F_f = mu_k*m*g
mu_k (coefficient of friction) = F_f/(m*g) = 3N/(6.0kg*9.81m/s^2) = 0.05096 ~ .051

[badboyben03]

Originally posted by deltabourne
edit: whoops, didn't see it

To keep something at constant velocity the net Force must be zero.. so here we have F_applied - F_friction = 0

or F_applied = F_friction = mu_k*m*g = .1*57*9.81 = 55.917

yeah i figured that one out later on, but thanks anyway

[badboyben03]

Originally posted by deltabourne
ok.. again, just set up your net f_x forces. We have the applied force (let's call it F_a) and the frictional force (F_f)

Net F_x = F_a - F_f = m*a
Solving for F_f, we have F_a - m*a = F_f

The applied force is 45N, the mass is 6.0kg, and acceleration is 7.0 m/s^2.. plug them in, you get a frictional force of 3N

Now, F_f = mu_k*m*g
mu_k (coefficient of friction) = F_f/(m*g) = 3N/(6.0kg*9.81m/s^2) = 0.05096 ~ .051



thanks again! do you have AIM?

[deltabourne]

Originally posted by badboyben03
thanks again! do you have AIM? yep: simplydelta

feel free to message me anytime