A sled of mass m = 22.0 kg is accelerated from rest on a frictionless, horizontal surface...

Question

A sled of mass m = 22.0 kg is accelerated from rest on a frictionless, horizontal surface to a velocity of v = 12.5 m/s, as it travels from x = 0 to x = 30 m. The graph displays a graph of force versus distance for the sled in terms of Fmax.

(a) Calculate the velocity of the sled at x = 100 m.

**Note: The graph appears in "images" if you search for this question. Please help.

Answer 1

Hi,

(Assuming I found the correct graph) I approached this problem using the work-energy theorem, the net work done to an object is equivalent to its change in kinetic energy. But first, we needed to determine some information.

Using the initial data of the object's motion (v_o = 0 m/s, v = 12.5 m/s, and x = 30 m), I used kinematics to determine the acceleration. (v² = v_o² + 2ax). The acceleration of the object is then 2.60 m/s².

I then used Newton's Second Law of Motion, an acceleration is proportional to the net force and inversely proportional to the mass, to determine F_max. (m = 22 kg, a = 2.60 m/s²). F = m·a. ·So F_max = 57.2 N. This means each interval on the y-axis is approximately 19.1 N.

Next, we can solve for the work done by the force by solving for the area under (or above when it's below the x-axis) the line. When the graph is flat, we can use W = F·d and when the graph has a slope, we can use W = 1/2 (F_o + F)·d. (The F_o and F are the initial and final force values for that segment in the graph.)

I calculated the following works for each segment.

W = 1716 J from 0-30m

W = 572 J from 30-50 m

W = -47.8 J from 50-55 m (approximating this distance, it may be to 56 m)

W = -669 J from 55 to 90 m

W = -95.5 J from 90-100 m

The net work done by the force is approximately 1480 J. Returning to the original approach, we can set this equal to the change in kinetic energy of the 22 kg object. KE = 1/2 m v². You can start with the initial velocity as 0 m/s at 0 m, so the initial KE is equal to 0 J. 1480 J = 1/2 m v².

The velocity is 11.6 m/s.

I know this is not the quickest approach, but I'm assuming the solution is using kinematics or dynamics, so I included a mix of approaches.

I hope this helps,