A 2.00-kg block is released from point A as shown in Figure 8.P26. The curved po

Question

A 2.00-kg block is released from point A as shown in Figure 8.P26. The curved portion of the track is frictionless The Sr zontal track, including the portion under the sprine has kinetic coefficient of friction mu_k = 0.100. (a) What is the block's velocity at point B? (b) What is the block's velocity at point C where the spnng's length is equal to its natural length? (c) How much docs the block compress the spring (k = 30.0 N/m) before coming momentarily to rest? (d) Where docs the block ultimately come to rest?

Explanation / Answer

here,
mass of block, m = 2 kg
height of path, h = 1.50m
coefficient of friction, uk = 0.100
distance to spring for point B, d = 10 m

Part A:
From Conservation of Energy
Change in Potential Energy = kinectic erngy at bottom,
mg*h = 0.5 * m * vb^2

Solving for Velocity at point B, vb
vb = sqrt(2*g*h)
vb = sqrt(2*9.8*1.50)
vb = 5.422 m/s

part B:
From newton Second law, Fnet = 0
F - Ff = 0
ma = uk*mg

solving for acceleration of block, a
a = uk*g
a = 0.1 *9.8
a = 0.98 m/s^2 -------------------------------(1)

From third eqn of motion, solving velocity at spring,
vs = sqrt(2*a*d)
vs = sqrt(2*0.98*10)
vs = 4.427 m/s

Part C:
From Conservation of Energy,
potential Energy gained by spring = loss of KE by block
0.5 * k * x^2 = 0.5 * m * vs^2

solving for compressed distance , x
x = sqrt(m*vs^2/k)
x = sqrt(2*4.427^2/30)
x = 1.143 m

Part D:
From Conservation of Energy,
loss of potential Energy = Kinetic Energy gained by block
0.5 * k * x^2 = 0.5 * m * v^2

solving for expulsion velocity, v
v = sqrt(k*x^2/m)
v = sqrt(30*1.143^2/2)
v = 4.427 m/s

then the track will become friction, so from1 a = 0.98 m/s

Solving for Distance to be stopped, d

d = v^2/2*a
d = 4.427^2/(2*0.98)
d = 10m

The object will stop at point B.

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