Question

If the instant t = 0, a force F = kt (k is a constant) acts on small body of mass m resting on a smooth horizontal plane. The time, when body leaves the surface is​

Answer 1

Answer:

mg/ksin α

Explanation:

mg=fsinα

as we know  f=kt

mg=ktsinα

mg/k sin alpha=t

Answer 2

The question is incomplete .

Question :

Choose the correct option :

If the instant t = 0, a force F = kt (k is a constant) acts on small body of mass m resting on a smooth horizontal plane. The time, when body leaves the surface is :

$a. \: mg \: k \: sin \alpha$

$b. \: k \: \div (mg \: sin \alpha )$

$c. \: (mg \: sin \alpha ) \div k$

$d. \: mg \div (k \: sin \: \alpha )$

Answer :

$The \: correct \: option \: is \: option \: d . \: mg \div (k \: sin \alpha )$

Referring to the given diagram :

m is the mass of the body

g is the acceleration due to gravity

$\alpha \: is \: the \: angle$

F or (f) is the force acting on the body

k or (k) is a constant

t is the time

According to the problem ,

$f \: sin \: \alpha = mg$

$therefore \: \\ kt \: sin \: \alpha = mg$

$t = mg \: \div (ksin \alpha )$

Hence, the time when body leaves the surface is

$mg \div (ksin \alpha )$

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