Question

An object of mass 1700 kg is moving at an initial velocity 11.3 x̂ x^ m/s . A force of -20.2 ŷ y^ kN is exerted on it for a time of 0.30 seconds. Calculate the final speed of the object in m/s to 3 sf.

Answer 1

Given that Initially object is traveling with velocity = V0x = (11.3 i) m/s

Now when a force of F = -20.2 kN in y-direction applied on the block, then

F = (-20200 j) N

From Newton's 2nd law:

F_net = m*a = -20200 N

a = F_net/m = -20200/1700 = -11.882 m/s^2 in y-direction

So, Using 1st kinematic equation in y-direction:

Vy = V0y + a*t

V0y = Initial vertical velocity = 0 m/s

t = time interval for which force is applied = 0.30 sec

So,

Vy = 0 - 11.882*0.30 = -3.5646 m/s

So after 0.30 sec, velocity of object will be:

V = Vx i + Vy j

Since there is no acceleration in x-direction, So velocity in x-direction will remain constant.

V = 11.3 i - 3.5646 j

Now speed of object will be:

|V| = sqrt (11.3^2 + (-3.5646)^2) = 11.849 m/s

In 3 significant figures

|V| = final speed = 11.8 m/s

Let me know if you've any query.