Question

a body of mass 5 kg is projected up a broad inclined plane at 30 degree to the horizontal with an initial velocity of 6 metre per second if the frictional force opposing its motion is 4.5 Newton find the distance it travels before coming to rest and its increase in potential energy at the end of the run​

Answer 1

initial velocity of body, u = 6m/s

final velocity of body, v = 0 m/s

mass of body ,m = 5kg

from Newton's 2nd law,

Fnet = mgsin30° + fr [ we know, frictional force acts just opposite the motion of body, so it acts downward along plane ]

or, ma = mgsin30° + 4.5N

or, 5a = 5 × 10 × 1/2 + 4.5

or, 5a = 29.5 N

or, a = 5.9m/s² ≈ 6 m/s²

now, using formula,

v² = u² + 2as

or, 0² = 6² + 2 × 6 s

or, s = -3m , here negative sign indicates that displacement is just opposite to the direction of acceleration.

so, distance travelled by body before coming to rest is 3m

from law of conservation of energy,

change in kinetic energy + workdone by frictional force = change in potential energy

or, 1/2 m(u² - 0²) + fr × s cos180° = ∆U

or, 1/2 × 5 × 6² + 4.5 × 3 × (-1) = ∆U

or, 90 - 13.5 = ∆U

or, ∆U = 76.5 J

hence, increase in potential energy is 76.5J

[ note : this question is not for 10th class student. so, you can't get some of concepts. for complete understanding. you have to learn 11th physics [chapter " friction", " work-power energy" and law of motion ]