Question

A block of mass m = 2.0 kg slides head on into a spring of spring constant k = 260 N/m. When the block stops, it has compressed the spring by 14 cm. The coefficient of kinetic friction between block and floor is 0.42. While the block is in contact with the spring and being brought to rest, what are

(a) the work done by the spring force and
(b) the increase in thermal energy of the block-floor system?
(c) What is the block's speed just as the block reaches the spring? Please show all steps.

Answer 1

a) Ws = 2.548 J

b) Wf = 1.153 J

c) v = 1.923 m / s

Step-by-step explanation:

a) The work done by the spring force

Ws = ½ * k * x²

Ws = ½ * 260 N/m *0.14² m

Ws = 2.548J

b) The increase in thermal energy can by find using

Et = Wf

Wf = µ * m *g * x

Wf = 0.42 * 2.0 kg *9.8 m/s² * 0.14m

Wf = 1.153 J

c) The speed just as the block reaches can by fin using

EK = Ws + Et

Ek = ( 2.548 + 1.153 ) J = 3.7 J

Ek = ½ * m * v²

v² = 2* Ek / m

v = √[2 * 3.7 J / 2.0 kg]

v = 1.923 m / s