A loaded sled weighing 1250 lb is given a speed of 2.5 mi/hr while moving a distance of 140 feet from rest on a horizontal surface. If the coefficient of friction is 0.105 what is the constant force applied horizontally?

Hint: calcularte the acceleration, and from that the sum of forces required to accelerate it can be found from \sum \vec F = m \vec a. The sum of forces here is the force applied minus the force of friction:

\sum \vec F = \vec F_{push} - \vec F {friction}

The negative sign is because friction opposes the movement of the sled.

So solve for \vec F_{push} - -what do you get?

Hello,

It seems that yo are the expert? Can you please answer this?

A 3,647 KG elevator is pulled upward by a constant tension force. The elevator starts at rest and under the influence of this constant tension force, accelerates upwards so that its speedis 6.82 m/s when it located 6.0 m above its initial location. What is the average power that the tension in the cable exerts on the elevator? Assume that the only force acting on the elevator is gravity. Write your answers in kW (kilo-watts) to three significant figures. (Hint: how long does it take for the situation to occur?
What’s the ratio of the instantaneous power exerted on the elevator by the tension in the cable when the elevator’s height is 2.74 m above its initial point tit he average power calculated in the question above? Write your answers to three significant figures( Hint you can use your calculations of the energy transferred to the elevator by the cable to determine the magnitude of the tension force)


Thank you!

chicas2013 - pease don't double-post questions. You posted this question here:
https://www.askmehelpdesk.com/physics/free-physics-answers-744301.html
And that's where you can expect a response.