Question

3. A block with a mass of 5kg is given a kick up a ramp angled at 35º such that its initial speed is 7.8 m/s. The block travels up the ramp for 4.3m before it stops moving and starts to slide back down. a) Find the initial mechanical energy (K + U) of the block. b) Find the final mechanical energy of the block. c) The answer to (a) is bigger than the answer to (b). That means the block lost energy to friction along the way. How much work did friction do against the block? d) Determine the force of friction in Newtons. e) Find the coefficient of friction Uk.

Answer 1

Given the mass of the block m = 5kg, the angle of inclination $\theta = 35\degree$ , the initial speed of the block u = 7.8m/s and the distance travelled by the block L = 4.3m.

(a) Since the initial height is zero, the initial potential energy is given by,

$U_{i} = mgh = 5\times 9.8\times 0=0J$

The initial kinetic energy K is given by,

K; = -mu? = x 5 (7.8)2 = 152.10J

The initial mechanical energy is given by,

TE K; +U; = 152.10 +0= 152.10J

So the initial mechanical energy is 152.10J.

(b) The block is fially at a height h. So the final potential energy of the block is given by,

Uf = mgh = mgL sin = 5 x 9.8 x 4.3 x sin 35º = 120.85J

Since the block comes to rest, the final speed of the block v = 0. So the final kinetic energy is given by,

$K_{f} = \frac{1}{2}mv^{2}=\frac{1}{2}\times 5\times 0^{2}=0J$

The final mechanical energy is given by,

So the final mechanical energy is 120.85J.

(c) Let Wf be the work done by friction. It is given by,

$W_{f}=TE_{i}-TE_{f}=152.10-120.85=31.25J$

So the work done by friction is 31.25J.

(d) The work done by friction f is given by,

$W_{f}=fL$

$f=\frac{W_{f}}{L}=\frac{31.25}{4.3}=7.27N$

So the force of friction is 7.27N.

(e) If $\mu_{k}$ is the coefficient of friction, then

$f=\mu_{k}mg$

$\mu_{k}=\frac{f}{mg}=\frac{7.27}{5\times 9.8}=0.15$

So the coefficient of friction is 0.15.