Question

In the figure, a 4.3 kg block is accelerated from rest by a compressed spring of spring constant 660 N/m. The block leaves the spring at the spring's relaxed length and then travels over a horizontal floor with a coefficient of kinetic friction μk = 0.260. The frictional force stops the block in distance D = 7.8 m. What are (a) the increase in the thermal energy of the block–floor system, (b) the maximum kinetic energy of the block, and (c) the original compression distance of the spring?

Answer 1

m*g = 4.3kg * 9.8N/kg = 42.14 N. = Wt. of block.

b. a = u*g = 0.26 * (-9.8) = -2.548 m/s^2

V2^2 = V1^2 - 2*2.548*7.8 = 0
V1^2 = 39.75
V1 = 6.30 m/s = Initial velocity.

KE = m*V1^2/2 = 42.14*39.75/2 = 838 J.

c. = (1m/660N) * 42.14N = 0.0638 m.

Answer 2

(a) W=F*X
W= u*Fn*X
=>M*g*uk*X -> 4.3kg*0.26*7.8m

(b) same as (a)

(c) Us=0.5*k*x^2
=> (a)=0.5*640N/m*x^2
figure out X from that


[very late but someone might benefit from this]