A 58.6-kg person, running horizontally with a velocity of +3.99 m/s, jumps onto

Question

A 58.6-kg person, running horizontally with a velocity of +3.99 m/s, jumps onto a 10.7-kg sled that is initially at rest. (I have the correct answer for part A, but i'm not sure what to do for part B)

(a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (Indicate the direction with the sign of your answer.)
*the answer for (a) is 3.37*

(b) The sled and person coast 31.1 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?

Explanation / Answer

let

M = 58.6 kg

u = 3.99 m/s

m = 10.7 kg

a) let V is the combined velocity

Apply conservation of momentum

M*u = (M + m)*V

==> V = M*u/(M+m)

= 58.6*3.99/(58.6+10.7)

= 3.37 m/s

b) acceleration of the sled, a = (vf^2 - vi^2)/(2*d)

= (0^2 - 3.37^2)/(2*31.1)

= -0.18258

we know, a = -g*mue_k

==> mue_k = -a/g

= -(-0.18258)/9.8

= 0.0186

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.