Question

Find the velocity of block at time t = 4 sec and t = 10 sec
A) 4m/S2, 2.5 m/s
B) Om/s, Om/s
C) Om/S, 2.5 m/s
D) 2.5m/s, 2.5m/s​

Answer 1

Given:

A 4 kg block is being pulled upwards with the variable force.

To find:

Velocity of the block at t = 4 sec and t = 10 sec.

Calculation:

The minimum time required for the force to lift the 4 kg block will be :

$\therefore \: F = mg$

$= > \: 5t= 4 \times 10$

$= > \: t= 4 \times 2$

$= > \: t= 8 \: sec$

So, till t = 8 seconds, the block will remain stationary.

Hence Velocity at t = 4 sec will be 0 m/s.

Now , we need to apply calculus to solve the remaining problem:

$\therefore \: a = \dfrac{5t}{4} - g$

$= > \: \dfrac{dv}{dt} = \dfrac{5t}{4} - g$

$\displaystyle = > \: \int dv = \dfrac{5}{4} \: \int t \: dt - g \int dt$

Putting limits:

$\displaystyle = > \: \int_{0}^{v} dv = \dfrac{5}{4} \: \int_{8}^{10} t\: dt - 10 \int_{8}^{10}dt$

$\displaystyle = > \: v= \frac{5}{8} \bigg \{ {t}^{2} \bigg \}_{8}^{10} - 10 (10 - 8)$

$\displaystyle = > \: v= \frac{5}{8} (100 - 64) - 20$

$\displaystyle = > \: v= ( \frac{5}{8} \times 36) - 20$

$\displaystyle = > \: v= 22.5 - 20$

$\displaystyle = > \: v= 2.5 \: m {s}^{ - 1}$

Velocity at t = 10 sec will be 2.5 m/s.

HOPE IT HELPS.

Answer 2

✿$\underline{\textbf{\textsf{ \purple{Solution}:- }}}$

First we need to find out the minimum time taken by the block to leave the surface.

When the block will leave the surface,

$\sf{F = mg}$

$\longrightarrow \sf{5t = 4 * 10}$

$\longrightarrow \sf{ {t}_{min} = 8 sec}$

$\sf{\\}$

$\underline{\bf{ \red{ {1}^{st}\:part } :- }}$

As the minimum time required by the body to leave the surface is 8 seconds, that means the body will still be at rest at t = 4 second. So, velocity at t = 4 sec will be,

$\longrightarrow \underline{\underline{\sf{ \green{ {v}_{4} = 0 m/s } }}}$

$\sf{\\}$

$\underline{\bf{ \red{ {2}^{nd}\:part } :- }}$

At t = 10 sec, the body will have left the surface and moving upwards with an acceleration.

Let the upward acceleration of the body be $\sf{a\:m/s^2}$

Then, from F.B.D we can say that,

$\sf{5t - mg = ma}$

$\longrightarrow \sf{a = \dfrac{5t - mg}{m}}$

$\longrightarrow \sf{a = \dfrac{5}{m}t - g}$

$\longrightarrow \sf{a = \dfrac{5}{4}t - g}$

$\longrightarrow \sf{ \dfrac{dv}{dt} = \dfrac{5}{4}t - g}$

$\longrightarrow \displaystyle \sf{ v = \int \dfrac{5}{4}t - g }$

Putting limits,

$\displaystyle \longrightarrow \sf{ v = \int_{8}^{10} \dfrac{5}{4}t - g }$

$\displaystyle \longrightarrow \sf{v = \int_{8}^{10} \dfrac{5}{4}t\:dt - \int_{8}^{10}g\: dt}$

$\longrightarrow \sf{v = \dfrac{5}{4}\Bigg\{\dfrac{t^2}{2}\Bigg\}_{8}^{10} - {{g}_{}}_{8}^{10} }$

$\longrightarrow \sf{v = \dfrac{5}{8}\Bigg\{10^2 - 8^2\Bigg\} - 10\Bigg\{10 - 8\Bigg\}}$

$\longrightarrow \sf{v = \dfrac{5}{8}36 - 20}$

$\longrightarrow \sf{v = 22.5 - 20}$

$\longrightarrow \underline{\underline{\sf{ \green{ {v}_{10} = 2.5\:m/s } }}}$

$\sf{\\}$

Hence, the answer is,

$\longrightarrow \underline{\underline{\sf{ \green{A)\:0m/s,\:2.5m/s} }}}$