Question

you apply a force of 10N on an initially stationary block of mass 3kg until ot raches a speed os 2m/s. for how long did hoh allmh the force? _s

2. tou travel 30 meters north followed by 40m west. what is the difference (in m) between your total distance traveled and the magnitude of your total displacement?

Answer 1

The force applied on the 3 kg block = 10 N

From Newtons's second law , acceleration = Force/mass

= 10/3 m/s²

Let for time 't' 10 N force is applied so that it reaches a speed of 2m/s .

For constant acceleration ,

v = u +at

v = velocity at time t

u = initial velocity (i.e. at t=0)

a= acceleration

t = time

Here final velocity became 2m/s therefore v = 2m/s .

u = 0 m/s ( initially the block is in rest)

a = 10/3 m/s²

Therefore ,

2 = 0 + 10/3*t

t = 0.6 seconds

Thus , 10 N force is applied on block for 0.6 seconds to reach the speed of 2m/s.

2 ) .

The distance is length of the path travelled by the object whereas the displacement is straight line joining initial and final point .

The distance in this case = 30 + 40 = 70 m (see fig.)

The displacement in this case can be calculated using the pythagoras theorem(see fig.) = = 50 m

Therefore , difference between distance and displacement = 70 - 50 = 20 m