Question

A box slides down a frictionless ramp.if it starts at rest, what is it’s speed at the bottom?

Answer 1

8.854 m/s is the speed of the box after it reaches bottom of the ramp.

Explanation:

From the figure we came to know that height of the block is 4 m.

We know that,

Total "initial energy of an object" = Total "final energy of an object "

Total "initial energy of an object" is = "sum of potential energy" and "kinetic energy" of an object at its initial position.

$\text { "g" acceleration due to gravity is } 9.8 \mathrm{m} / \mathrm{s}^{2}$

$\text { Total initial energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{i}}+\frac{1}{2} \mathrm{m} v_{i}^{2}$

Initial velocity is “0” as the object does not have starting speed

$\text { Height of the block where the object is placed initially }\left(h_{i}\right) \text { is } 4 \mathrm{m} \text { . }$

$\text { Total initial energy }=\mathrm{m} \times 9.8 \times 4+\frac{1}{2} \mathrm{m} 0^{2}$

Total initial energy = 39.2 × m

$\text { Total final energy }=\mathrm{m} \times \mathrm{g} \times \mathrm{h}_{\mathrm{f}}+\frac{1}{2} m v_{f}^{2}$

$\text { We need to find final velocity } v_f$

$\text { Height of the block where the object is travelled to bottom (h_) is } 0 \mathrm{m} \text { . }$

$\text { Total final energy }=\mathrm{m} \times 9.8 \times 0+\frac{1}{2} m v_{f}^{2}$

Now,  Total initial energy of an object = Total final energy of an object

$39.2 \times \mathrm{m}=0.5 \mathrm{m} v_{f}^{2}$

$\frac{39.2}{0.5}=v_{f}^{2}$

$v_{f}^{2}=78.4$

$v_{f}=\sqrt{78.4}$

$v_{f}=8.854 \mathrm{m} / \mathrm{s}$

Final speed is 8.854 m/s.