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

To find the speed of the block at the instant of collision with the spring, we can use the conservation of mechanical energy principle.

Step-by-step explanation:

To find the speed of the block at the instant of collision with the spring, we can use the conservation of mechanical energy principle. Initially, the block has kinetic energy due to its motion, which is converted into potential energy when the block compresses the spring. The work done by friction dissipates some of this mechanical energy. So, we can set up the equation:

K_initial - Work_friction = U_spring

where K_initial is the initial kinetic energy, Work_friction is the work done by friction, and U_spring is the potential energy stored in the spring. Since we know the mass of the block, the distance the spring is compressed, and the force constant of the spring, we can calculate the speed of the block at the instant of collision with the spring.

Answer 2

`v=65m/s`

Step-by-step explanation:

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

`E_1=(1)/(2)*K*x^2`

`x=4.16cm`

`E_1=(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=(1)/(2)*m*v^2`

Solve to v' to find the velocity

`v=\sqrt{(2*K_E)/(m)}`

`v=\sqrt{(2*0.283J)/(1.34kg)}=√(0.424 m^2/s^2)`

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

`v=65m/s`