1. Physics Learning Objects 1
1. A block (0.2 kg) attached to a horizontal spring is stretched d=20cm and released.
The released block goes through the black shaded area which has a friction factor
μ=0.3, climbs up to an inclination with a smooth arc and comes back. If the black
shaded area is S meters long, where will the block stop? k=10N/m, g=9.8N/kg,
S=20cm.
Explanation:
The total energy of block at initial status is stored as potential energy of the spring
(E=U=1/2kA^2). When the block is released, the potential energy transformed to
kinetic energy, potential energy and the work done by the block against friction force.
At last, the block stops. Since the kinetic energy of the stopped block is zero and the
potential energy of spring is zero, all the energy is transformed to the work done by
the block against friction force (W=F*S=μ*m*g*A). A supposed to be all the
frictional distance the block travelled.
Therefore, we get:
1
2
𝑘𝑑2
= 𝜇𝑚𝑔𝐴
1
2
∗
10𝑁
𝑘𝑔
∗ 0.2𝑚2
= 0.3 ∗ 0.2kg ∗
9.8N
kg
∗ A
A = 0.34m
A/S=0.34m/0.2m=1.7
2S-A=0.06m
The block stops at 6cm from the start of the black shaded area (from left to right).