# A 3 kg block travels along a horizontal surface with a coefficient of kinetic friction of 0.26 at...

## Question:

A 3 kg block travels along a horizontal surface with a coefficient of kinetic friction of 0.26 at a speed of 7 m/s. After sliding a distance of 1.0 m the block makes a smooth transition to a ramp with a coefficient of kinetic friction of 0.26. How far up the ramp does the block travel before coming to a momentary stop?

## Frictional energy:

The friction force acts in the opposite direction of the motion, therefore the work done by the friction is decreasing the energy of the object. Or we can say the object has to spend the energy to overcome the friction.

## Answer and Explanation:

Given data:

• mass of the block {eq}\rm (m) = 3 \ kg {/eq}
• Initial velocity of the block {eq}\rm (u) = 7 \ m/s {/eq}
• Angle of ramp {eq}\rm (\theta) = 30^\circ \ [\text{Assumed, missing in question } ] {/eq}

Now, the initial energy of the block {eq}\rm E = \dfrac{1}{2} mu^{2} \\ E = \dfrac{1}{2} \times 3 \times 7^{2} \\ E = 73.5 \ J {/eq}

Now, energy lost in the friciton on horizontal plane would be

{eq}\rm W_{f1} = f_{1} \times d \\ W_{f1} = \mu mg \times d \\ W_{f1} = 0.26 \times 3 \times 9.8 \times 1 \\ W_{f1} = 7.644 \ m {/eq}

Now, let us consider that the block has travel "S m "along the incline

therefore, the energy lost in the friction and the increase in the potential energy would be

{eq}\rm E_{2} = f \times S + mg(S\sin\theta) \\ E_{2} = (\mu \times mg \times S) + (mg S \sin\theta) \\ E_{2} = (0.26 \times 3 \times 9.8 \times S ) + (3 \times 9.8 S \sin30^\circ) \\ E_{2} = 22.34S {/eq}

now, applying the energy conservation

{eq}\rm E - W_{f1} = E_{2} \\ 73.5 -7.644 = 22.34 S \\ S = 2.95 \ m {/eq}