Question

A 6 kg ball starts from rest and rolls down a rough gradual
slope until it reaches a point 80 cm lower than its starting
point. Then the speed of the ball is :​

Answer 1

Answer:

V=14^0.5

Explanation:

Its pe has converted to ke as

if we consider potential at 80 cm below as reference =0

then pe at 80 cm above will be 60*0.8 =42

then it has converted to 3*v^2=42

v=14^0.5

Answer 2

Answer:

The speed of the ball is 4 m/s.

Explanation:

We will solve this question through the energy conservation principle.i.e.

$\frac{1}{2}mv^2-\frac{1}{2}mu^2=mgh$           (1)

Where,

m=mass of the body

v=final velocity of the body

u=initial velocity of the body

h=height the covers

g=acceleration due to gravity=10 ms⁻²

According to the question,

m=6kg

h=80cm=0.8m

u=0 (as the ball starts from rest)

By putting these values in equation (1) we will get the speed of the ball,

$\frac{1}{2}mv^2-\frac{1}{2}m(0)^2=mgh$

$\frac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$     (2)

Now, let's substitute the value of g and h in equation (2) we get;

$v=\sqrt{2\times 10\times0.8}$

$v=4m/s$

Hence, the speed of the ball is 4 m/s.