A 61.5 kg person, running horizontally with a velocity of +4.00 m/s, jumps onto a 13.0 kg sled ---?

Question

--that is initally at rest.

(b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?

Answer 1

This is a balance of forces problem invoking Newton's laws of motion.

Here we have mV = (m + M)v; which is Newt's conservation of momentum and m = 61.5 kg, M = 13 kg, and V = 4 mps. Solve for v, the initial velocity of the person-sled combo.

Now invoke the SUVAT equation, v^2 = 2aS; where S = 30 meters and v is the combo initial velocity found above. Solve for a = f/(m + M) = v^2/2S; where f is the net force acting on the combo as it decelerates at a. f = (m + M)a is the force bringing the combo to a halt.

The source of f is the friction acting on the sled. Thus, the net force f = (m + M)a = kN = k(m + M)g, where N is the normal weight of the combo = (m + M)g and g = 9.81 m/sec^2 on Earth's surface.

Now you can write (m + M)a = k(m + M)g; so that a = v^2/2S = kg and k = v^2/2gS the coefficient of friction in the snow. v, g, and S are given; so you can do the math.

Some interesting physics to note in the k equation. If you doubled the initial velocity v of the combo, k would have to be four times greater than the original value because of the v^2 factor. And doubling V, the person's velocity before sled, would double the v. Another way to put this, if the person jumps on at 8 mps (2X4 mps), the sled would slide 120 m (4X30 m) if k were the same. This also works for doubling the speed of a car, that would take four times the distance to stop for a given friction force.

Answer 2

The first step is using conservation of momentum to determine the speed of the combined sled and person

61.5*4=74.5*v
v=61.5*4/74.5

then, look at the work done by friction
f*d=KE

f*30=.5*74.5*v^2
or
f=.5*61.5^2*16/(74.5*30)

then coefficient is
N*u=f
where N is the normal force:
74.5*9.81

u=0.019

j

Answer 3

a) the blended velocity of sled and guy could be 3.634m/sec b) the adverse acc experienced via the sled and guy combine is .22m/sec2 & I rather have forgotten the formula for coeff of Friction.