Question

A 2.5 kg block starts with a speed of 3 m/s, goes down a 4 m hill, and then slides up a new hill but stops when it is only 1.5 m high.
a) How much work was done by friction?
b) If there was no friction, how high could the block have slidden without stopping?

Answer 1

Answer:

See the answers below

Explanation:

This problem can be solved by means of the law of conservation of energy, which tells us that the energy between two points is conserved, that is, remains the same.

That is to say, we have two points, the point A where it starts to move and reaches the point B that is 1.5 [m] high, where it stops.

The key to solving this problem is to identify the types of energies at each point. At point A we have kinetic energy and potential energy where the block moves at speed 3 [m/s] at a height of 4 [m]. Whereas at Point B we only have potential energy, since the body is at a height of 1.5 [m], relative to the ground.

a)

Since between the displacement between points A and B there is a friction force, this friction force decreases the final energy in B, in this way the energy or work of the friction will have a negative sign.

$E_{A}-W_{A-B}=E_{B}$

Now replacing in the above equation.

$(\frac{1}{2}*m*v^{2}+m*g*h_{1})-W_{1-2}=m*g*h\\(\frac{1}{2}*2.5*(3)^{2}+2.5*9.81*4)-W_{1-2}=2.5*9.81*1.5\\\\W_{1-2}=72.56[J]$

b)

We must use the same equation, but this time eliminating the Working term due to friction.

$E_{A}=E_{B}\\\frac{1}{2}*m*v^{2}+m*g*h_{A}=\ m*g*h_{B}$

$(\frac{1}{2}*2.5*(3)^{2}+2.5*9.81*4) =\ 2.5*9.81*h_{B}\\h_{B}=4.45 [m]$

As we can see without friction the block can reach a higher height