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.
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?
2 answers
(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]
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]