SATHEE: Laws Of Motion Question 318

Laws Of Motion Question 318

Question: A body of mass 5 kg under the action of constant force $\vec{F} = F_{x} \hat{i} + F_{y} \hat{j}$ has velocity at $t = 0 s$ as $\vec{v} = (6 \hat{i} - 2 \hat{j}) m / s$ and at $t = 10 s$ gas $\vec{v} = 6 \hat{j} m/s$ . The force F is:

Options:

A) $(- 3 \hat{i} + 4 \hat{j}) N$

B) $(- \frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}) N$

C) $(3 \hat{i} - 4 \hat{j}) N$

D) $(\frac{3}{5} \hat{i} - \frac{4}{5} \hat{j}) N$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] From question, Mass of body, m = 5 kg Velocity at $t = 0, u = (6 \hat{i} - 2 \hat{j}) m / s$ Velocity at $t = 10 s, v = + 6 \hat{j} m / s$

Force, F=? $a = \frac{v - u}{t} = \frac{6 \hat{j} - (6 \hat{i} - 2 \hat{j})}{10} = \frac{- 3 \hat{i} + 4 \hat{j}}{5} m/ s^{2}$

Force, $F = m a = 5 \times \frac{(- 3 \hat{i} + 4 \hat{j})}{5} = (- 3 \hat{i} + 4 \hat{j}) N$

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Laws Of Motion Question 318

Question: A body of mass 5 kg under the action of constant force F→=Fxi^+Fyj^ has velocity at t=0s as v→=(6i^−2j^)m/s and at t=10s gasv→=6j^ m/s . The force F is:

Options:

Answer:

Solution:

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Question: A body of mass 5 kg under the action of constant force $\vec{F} = F_{x} \hat{i} + F_{y} \hat{j}$ has velocity at $t = 0 s$ as $\vec{v} = (6 \hat{i} - 2 \hat{j}) m / s$ and at $t = 10 s$ gas $\vec{v} = 6 \hat{j} m/s$ . The force F is: