Question

A spring (k = 190 N/m) is fixed at the top of a frictionless plane inclined at angle θ = 33 °. A 1.0 kg block is projected up the plane, from an initial position that is distance d = 0.60 m from the end of the relaxed spring, with an initial kinetic energy of 27 J. (a) What is the kinetic energy of the block at the instant it has compressed the spring 0.10 m? (b) With what kinetic energy must the block be projected up the plane if it is to stop momentarily when it has compressed the spring by 0.30 m?

Answer 1

Answer:

(a)$KE_{f}=24.30996 J$

(b)$KE_{f}=13.36 J$

Explanation:

(a)

Let compressed length be \triangle x

Initial distance, d= 0.6m

From the law of conservation of energy

$KE_{i}=KE_{f}+PE_{spring}+PE_{mass}$ where $KE_{i}$ is the initial kinetic energy, $KE_{f}$ is final kinetic energy, $PE_{spring}$ is potential energy of spring, $PE_{mass}$ is potential energy of the block

$27J=KE_{f}+0.5k(\triangle x)^{2}+mgh$ where m is mass, g is gravitational constant and h is height, k is spring constant=190

$KE_{f}=27-0.5k(\triangle x)^{2}-mg(d+\triangle x)sin 33^{o}$

$KE_{F}=27-0.5*190*0.1^{2}-(1*9.81*(0.60+0.10)sin 33^{o}$  

$KE_{f}=27- 0.95- 3.740036=24.30996 J$

$KE_{f}=24.30996 J$

(b)

When at rest, the final velocity is zero  

From the law of conservation of energy

$KE_{i}=KE_{f}+PE_{spring}+PE_{mass}$

$KE_{i}=0+0.5k(\triangle x)^{2}+mgh$

$KE_{f}=0.5k(\triangle x)^{2}+mg(d+\triangle x)sin 33^{o}$

$KE_{f}=(0.5*190*0.3^{2})+(1*9.81*(0.6+0.3)sin33^{o}$

$KE_{f}= 8.55+ 4.808618= 13.358618$

$KE_{f}=13.36 J$