Question

A luge and its rider, with a total mass of 91 kg, emerge from a downhill track onto a horizontal straight track with an initial speed of 27 m/s. If a force slows them to a stop at a constant rate of 4.4 m/s2, (a) what magnitude F is required for the force, (b) what distance d do they travel while slowing, and (c) what work W is done on them by the force? What are (d) F, (e) d, and (f) W if they, instead, slow at 8.8 m/s2?

Answer 1

(a) F =170 N

(b) d = 342.25 m

(c) W= -58182.5 J

(d) F=340 N

(e) d = 171.125 m

f) W= -58182.5 J

Step-by-step explanation:

Given

Total mass of the luge and the rider, m = 85 kg.

lnitial speod, Vo= 37m/s.

Acceleration, a =-2 m/s^2, (Negative sign because the force slowing the

luge and its rider to zero speed).

Final speed vf = 0 m/s.

2). Force = mass acceleration (Newton's Second Law of motion.)

Magnitude of the force= m/a/

= 85 x /- 2/

= 85 x 2

= 170N

3). Using kinematic Equations to describe the motion of the luge and the rider:

=V + 2ad...(2)

Where d is the distance they travel in meters while slowing dowm (required)

V 1s the linal spced and equals 0 m/s in this problem, v, 1s the initial speed

and equals 37 m/s in this problem, and a is the acceleration and equals -2

m/s* in this problem. Substitute in equation in 2: 0= 37+2 x (-2d)

4d= 1369

d= = 342.25 m.

4). W= F.d= /F//d/cos Theta... (3)

Where 6 is the angle between the torce and the displacement vectors.

There is no need to use the integral form to find the work done

becalise the torce is const ant and not a function of the displacement.

As the luge and its rider emerge from a downhill track onto a horizontal

straight track and the force slows them to stop, then the force opposes the

direction of the displacement. The two vectors (force and displacement) in

this case are in opposite directious, which means the angle 6 = 180°.

Substituting in to equaton 3:

Work done on the luge and its rider by the force = W

W = 170 x 312.25 x cos 180° = 170 x 312.25 x -1 = -58182.5 J.

5). If the acceleration=-4 m/s instead of -2 m/s

Substitute in equation 4 by the new value of the acceleration to get the

magnitude of the force acting on the luge and the rider as follows:

Magnitude of the force = F

F= 85 x/-4/

F= 85x 4

=340 N.

6). Substitute into equation 2 by the new value ot the acceleration to get the

distance they travel:

0= 37^2 + 2 x(-4d)

8d=1369

d= 1369/8

Therefore, d= 171.125 m.

7). Substitute in equation 3 by the new values of the force and the displacement,

given m part d and e, to get the work done on them (Theta = 180 as explained earlier):

W=340 x 171.125 x -1 = -58182.5 J.