Question

(a) A body of mass 10 kg is initially moving with a speed of 4.0m s

–1

. A force of 30 N is now

applied on the body for 2 seconds.

(i) What is the final speed of the body after 2 seconds?

(ii) How much work has been done during this period?

(iii) What is the initial kinetic energy?

(iv) What is the final kinetic energy?

(v) What is the distance covered during this period?

(vi) Show that the work done is equal to the change in kinetic energy?​

Answer 1

Given :-

Mass of body = m = 10 Kg

Initial Velocity of body = u = 4 ms-¹

Force acting = F = 30 N

Time = t = 2 s

Here we will find acceleration of the body.

F = ma

30 = 10a

a = 3 ms-²

Using first equation of motion we will find the final Velocity of body.

i). v = u + at

v = 4 + 3 × 2

v = 10 ms-¹

Again, using third equation of motion we will find the distance travelled.

Let the distance travelled = S

v² = u² + 2aS

(10)² - (4)² = 2 × 3 × S

6S = 84

S = 14 m

ii). For work done,

W = FS

W = 30 × 14

W = 420 J

iii). For initial Kinetic energy,

K.E. = 1/2 mu²

K.E. = 1/2 × 10 × 4 × 4

K.E. = 80 J

iv). For Final Kinetic Energy,

K.E.' = 1/2 mv²

K.E.' = 1/2 × 10 × 10 × 10

K.E.' = 500 J

v). As calculated above distance travelled,

S = 14 m

vi). W = ∆K.E.

W = K.E.' - K.E.

W = 500 J - 80 J

W = 420 J