Question

A small steel ball falls through a syrup at a constant speed of 10 m/s. If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upwards?

Answer 1

Hello dear,

● Answer -

u = 14.14 m/s

● Explaination-

Ball falls with constant velocity v.

v^2 = u^2 + 2gh

10^2 = 0 + 2×10×h

h = 5 m

When it's thrown upwards, force is twice its weight.

Hence, acceleration is twice the gravity in opposite direction.

a = -2g

Using Newton's 3rd law of kinematics,

v^2 = u^2 + 2(-2g)h

0 = u^2 - 4×10×5

u^2 = 200

u = 14.14 m/s.

Therefore, the ball will move upwards with 14.14 m/s.

Hope this helps...

Answer 2

Answer:

10 m/s

Explanation:

When the ball is falling down, force of buoyancy and weight balance each other. as it moves at constant speed.

If the ball is pulled at double its effective weight, then bouyant force and weight will together balance it and it moves at constant speed of 10m/s