Question

The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of friction between each block and the surface is 0.4 . The two blocks collide, stick together, and move to the right. Remember that the spring is not attached to the 8 kg block. Find the speed of the 8 kg block just before it collides with the 2 kg block. Answer in units of m/s.

Answer 1

The speed of the 8 kg block before the collision is found by considering the work done by friction and the initial potential energy in the spring and using the conservation of energy principle to solve for the speed.

Step-by-step explanation:

To find the speed of the 8 kg block just before it collides with the 2 kg block, we must calculate the motion of the block under the influence of friction before the collision occurs.

First, we determine the force of friction using the coefficient of friction (μ) and the normal force (N), which in this case equals the weight of the block (m × g) since the surface is horizontal. So, the force of friction is given by Fₙ = μ × m × g = 0.4 × 8 kg × 9.8 m/s² = 31.36 N.

Since there is friction, work done by friction must be considered to find the velocity. The work done by friction (Wₙ) is the force of friction (Fₙ) multiplied by the distance the block moves before colliding with the second block (d), which is not given in this problem. But if we had the distance, we could use the work-energy principle, where the net work done on an object is equal to the change in its kinetic energy (KE).

Assuming no other forces act on the block, then the work done by friction is equal to the initial potential energy stored in the spring. Using Hooke's Law, the potential energy (PE) in the spring is given by PE = ½ kx², where k is the spring constant and x is the compression distance. This energy gets converted entirely into the kinetic energy of the 8 kg block minus the losses due to friction.

Using the conservation of energy:

PE - Wₙ = ½ m v²

½ kx² - Fₙ × d = ½ m v²

We solve this equation for v, the speed of the block, if we know the distance (d).

Answer 2

3.258 m/s

Step-by-step explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

`\mu` = Coefficient of friction = 0.4

Energy stored in spring is given by

`U=(1)/(2)kx^2\\\Rightarrow U=(1)/(2)* 263* 0.7^2\\\Rightarrow U=64.435\ J`

As the energy in the system is conserved we have

`(1)/(2)mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2(U-\mu mgx)/(m)}\\\Rightarrow v=\sqrt{2(64.435-0.4* 8* 9.81* 0.7)/(8)}\\\Rightarrow v=3.258\ m/s`

The speed of the 8 kg block just before collision is 3.258 m/s