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Chapter 3: Problem 65

A constant force acting on a body of mass \(3.0 \mathrm{~kg}\) changes its speed from \(2.0 \mathrm{~ms}^{-1}\) to \(3.5 \mathrm{~ms}^{-1}\) in \(25 \mathrm{~s}\). The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force? (A) \(0.18 \mathrm{~N}\) (B) \(0.36 \mathrm{~N}\) (C) \(0.9 \mathrm{~N}\) (D) None of these

Short Answer

Expert verified
The magnitude of the force is \(0.18 \mathrm{~N}\) and the direction is the same as that of the initial velocity. The correct answer is (A).

Step by step solution

01

Find acceleration

We need to find acceleration (a) first, which can be calculated by subtracting the initial velocity (u) from the final velocity (v) and dividing the result by the time (t) taken for the change: \[a = \frac{v - u}{t}\] Using the given values in the problem, we have: \[a = \frac{3.5 \mathrm{~ms}^{-1} - 2.0 \mathrm{~ms}^{-1}}{25 \mathrm{~s}}\]
02

Calculate acceleration

Now, we calculate the acceleration: \[a = \frac{1.5 \mathrm{~ms}^{-1}}{25 \mathrm{~s}}\] \[a = 0.06 \mathrm{~ms}^{-2}\]
03

Apply Newton's second law

Now, we can use Newton's second law of motion to find the force (F) acting on the object, using the mass (m) and the acceleration (a): \[F = m \cdot a\] Plugging in the given mass and the calculated acceleration: \[F = 3.0 \mathrm{~kg} \cdot 0.06 \mathrm{~ms}^{-2}\]
04

Calculate the force

Finally, we calculate the force: \[F = 0.18 \mathrm{~N}\]
05

Determine the direction of the force

Since the direction of motion of the body remains unchanged, the force must be acting in the same direction as the initial velocity. Therefore, the magnitude of the force is \(0.18 \mathrm{~N}\) and the direction is the same as that of the initial velocity. So the correct answer is (A).

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Most popular questions from this chapter

A block of mass \(M\) is pulled along a horizontal frictionless surface by a rope of mass \(m\). If a force \(P\) is applied at the free end of the rope, the force exerted by the rope on the block is (A) \(\frac{P m}{M+m}\) (B) \(\frac{P m}{M-m}\) (C) \(P\) (D) \(\frac{P M}{M+m}\)

A body of mass \(2 \mathrm{~kg}\) moves vertically downwards with an acceleration \(a=19.6 \mathrm{~m} / \mathrm{s}^{2}\). The force acting on the body simultaneously with the force of gravity is \(\left(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right.\), neglect air resistance) (A) \(19.6 \mathrm{~N}\) (B) \(19.2 \mathrm{~N}\) (C) \(59.2 \mathrm{~N}\) (D) \(58.8 \mathrm{~N}\)

A girl of mass \(50 \mathrm{~kg}\) stands on a measuring scale in a lift. At an instant, it is detected that the reading reduces to \(40 \mathrm{~kg}\) for a while and then returns to original value. It can be said that (A) The lift was in constant motion upwards (B) The lift was in constant motion downwards (C) The lift was suddenly started in downward motion (D) The lift was suddenly started in upward motion

A pebble of mass \(0.05 \mathrm{~kg}\) is thrown vertically upwards. The direction and magnitude of the net force on the pebble is given below, choose the incorrect option. (A) During its upward motion, force is \(0.5 \mathrm{~N}\) in vertically upward. (B) During its downward motion, force is \(0.5 \mathrm{~N}\) in vertically downward. (C) At the highest point, where it is momentarily at rest, force is \(0.5 \mathrm{~N}\) in vertically downward. (D) If the pebble was thrown at an angle of say \(45^{\circ}\) with the horizontal direction, force is \(0.5 \mathrm{~N}\) in vertically downward (Ignoring air resistance).

Two masses \(m\) and \(M\) are connected by a light string passing over a smooth pulley. When set free, \(m\) moves up by \(1.4 \mathrm{~m}\) in \(2 \mathrm{~s}\). The ratio \(\frac{m}{M}\) is \(\left(g=9.8 \mathrm{~ms}^{-2}\right.\) ) (A) \(\frac{13}{15}\) (B) \(\frac{15}{13}\) (C) \(\frac{9}{7}\) (D) \(\frac{7}{9}\)
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