A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction, how fast will the ball be rolling when it reaches the bottom of the incline?

1.6s

Explanation:

Given that A 1.20 kg solid ball of radius 40 cm rolls down a 5.20 m long incline of 25 degrees. Ignoring any loss due to friction,

To know how fast the ball will roll when it reaches the bottom of the incline, we need to calculate the acceleration at which it is rolling.

Since the frictional force is negligible, at the top of the incline plane, the potential energy = mgh

Where h = 5.2sin25

h = 2.2 m

P.E = 1.2 × 9.8 × 2.2

P.E = 25.84 j

At the bottom, K.E = P.E

1/2mv^2 = 25.84

Substitutes mass into the formula

1.2 × V^2 = 51.69

V^2 = 51.69/1.2

V^2 = 43.07

V = 6.56 m/s

Using the third equation of motion

V^2 = U^2 + 2as

Since the object started from rest,

U = 0

6.56^2 = 2 × a × 5.2

43.07 = 10.4a

a = 43.07/10.4

a = 4.14 m/s^2

Using the first equation of motion,

V = U + at

Where U = 0

6.56 = 4.14t

t = 6.56/4.14

t = 1.58s

Therefore, the time the ball rolls when it reaches the bottom of the incline is approximately 1.6s

v 154.9= 0.9 length length s little over 154.9cm