Question

a 1.34kg block sliding on a horizontal surface collides with a spring of force constant 1.93 N/cm. The block compresses the spring 4.16cm from the extended position. Friction between the block and the surface dissipates 117mJ of mechanical energy as the block is brought to rest. Find the speed of the block at the instant of collision with the spring

Answer 1

Answer:

$v=65m/s$

Explanation:

Friction force to find the speed of the block at the instant of collision so

$E_1=\frac{1}{2}*K*x^2$

$x=4.16cm$

$E_1=\frac{1}{2}*1.93N/cm*(4.16cm)^2$

$E_1=0.167 J$

$E_2=117mJ=0.117J$

$K_E=E_1+E_2=284J$

$K_E=\frac{1}{2}*m*v^2$

Solve to v' to find the velocity

$v=\sqrt{\frac{2*K_E}{m}}$

$v=\sqrt{\frac{2*0.283J}{1.34kg}}=\sqrt{0.424 m^2/s^2}$

The speed of the block at the instant of collision with the spring is

$v=65m/s$