A 10.0kg stone slides down a snow-covered hill (the figure (Figure 1) ), leaving

Question

A 10.0kg stone slides down a snow-covered hill (the figure (Figure 1) ), leaving point A with a speed of 12.0m/s . There is no friction on the hill between points A and B , but there is friction on the level ground at the bottom of the hill, between B and the wall. After entering the rough horizontal region, the stone travels 100 m and then runs into a very long, light spring with force constant 2.30N/m . The coefficients of kinetic and static friction between the stone and the horizontal ground are 0.20 and 0.80, respectively. What is the speed of the stone when it reaches point B ? How far will the stone compress the spring? Will the stone move again after it has been stopped by the spring?

Explanation / Answer

By conservation of energy principle:

E=0.5*(10)*(12)^2+10*9.8*20=2680 J

This is the kinetic energy of the block

So

V^2=2*E/m=536

V=23.15 m/s

E= work done by friction + spring PE

2680=0.2*10*9.8(100+x)+2.3*0.5*x^2

2680=1960+19.6x+1.15x^2

Solving this x=17.911 m

C)

The block will move if the friction force is overcome by the spring force

That is if k*x>(mus)(m*g)

kx=17.9*2.3=41.19 N

(mus)(m*g)=78.4

Since the spring force is less than the maximum sustainable friction the block wont move.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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