Suppose you try to move a carate by tying a rope qround it and pulling on the rope at angle of $30^0$ above the horizontal. What is the tension required to keep the crate moving with constant velocity? Assume weight of the crate $W = 500N$ and coefficient of dynamic fraction $\mu_k = 0.40$.

Given,

Let $W$ be the weight of the crate, and $R$ be the normal reaction, $T$ be the tension and $f_s$ be the frictional force. 

Let $\theta = 30^0$ be the angle between $T$ and horizontal. Then,

W = R

In horizontal direction, $F = T\;cos\theta \; - \; F_s$

For constant velocity, $a = 0$

$T\cos\theta = F_s$

$T$ = $\frac{\mu_k\;*\;R}{cos\theta}$ = $\frac{0.4 \;*\; 500}{cos(30^0)}$ = $230.94 \; N$

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A block is pushed 1.5 m along a horizontal tabletop with horizontal force of 2.4 N. If the frictional force between the surface in contact is 0.6 N, what is (i) work done on the block by the frictional force and (ii) the total work done on the block?

Given,

Distance (d) = 1.5 N

Horizontal force ($F_H$) = 2.4 N

Frictional Force ($F_C$) = 0.6 N

(i) Work done by the frictional force ($W_f$) = $F_c * d$ = $-0.6 * 1.5$ = $-0.9 \; Joule$

(ii) Work done by the horizontal force ($W_H$) = $F_H * d$ = $2.4 * 1.5$ = $3.6 \; Joule$

Now, 

Total work done ($W$) = $W_f$ + $W_H$ = $3.6 - 0.9$ = $2.7 \; Joule$

  

In a physics lab experiment, a $6\;kg$ box is pushed across a flat table by a horizontal force $(F)$. (i) If the box is moving at a constant speed of $0.30\;m/s$ and the coefficient of kinetic friction is $0.12$, What is the magnitude of $F$? (ii) If the box is speeding up with a constant acceleration of $0.18\;m/s^2$, what will be the magnitude of $F$?

Given,

Mass of box $(m) = 6\;kg$

(i) Speed of box $(u) = 0.35\;m/s$

Coefficient of kinetic friction $(\mu_k) = 0.12$

Frictional Force $(F) = \;?$

We have,

$\mu_k = $$\frac{F_k}{R}$     ⇒   $F_k = \mu_k\;R = \mu_k\;m.g$

Now,

$F_u = F + F_k$                                       [∵ for Uniform Motion, $a = 0$]

$F_u = m.a + F_k = \mu_k\;m.g = 0.12\;*\; 6\;*\;10  = 7.2\;N$

(ii) When the box moves with acceleration $0.18\;m/s^2$. Then

$F_u = m.a + m_k\;m/g = 6\;*\;0.18 + 0.12\;*\;6\;*10 = 8.28\;N$

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A little red wagon with mass $7\;kg$ moves in a straight line on a frictionless horizontal surface. It has an initial speed of $4\;m/s$ and then is pushed $3\;m$ in the direction of the initial velocity by a force of $10\;N$. Calculate the wagon's final speed and the acceleration produced by the force.

Given,

Mass of wagon $(m) = 7\;kg$

Initial Speed $(u) = 4\;m/s$

Distance $(s) = 3\;m$

Force $(F) = 10\;N$

Final Speed $(v) = \;?$

Acceleration $(a) = \;?$

We have,

$F = ma$     ⇒    $a = $$\frac{10}{7}$  = $1.43\;m/s^2$

Again,

$v^2 = u^2 + 2\;a\;s = 4^2 + 2\;*\;1.43\;*\;4 = 27.44$        ⇒       $v = 5.2\;m/s$

ஃ Final speed is $5.2\;m/s$ & Acceleration is $1.43\;m/s^2$

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A ball of mass $0.05\;kg$ strikes a smooth wall normally four times in $2$ seconds with a velocity of $10\;m/s$. Each time the ball rebounds with the same speed of $10\;m/s$. Calculate the average force on the wall.

Given,

Mass of ball $0.05\;kg$

Initial Velocity $(u) = 10\;m/s$

Final Velocity $(v) = (-10\;m/s)$  [Opposite in direction]

No. of striking $(n) = 4$ times

Time taken $(t) = 2\;sec$

Average force on wall $(F) = \;?$

We have,

$F = $$\frac{n(mv-mu)}{t}$    =    $\frac{4(10\;*\;0.05 - (-10 \;*\;0.05))}{2}$    =    $2\;N$

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A $550\;N$ physics student stands on a bathroom scale in an elevator. As the elevator starts moving the scale reads $450\;N$. Draw a free body diagram of the problem and find the magnitude and direction of the acceleration of the elevator.

Given,

Weight of the body $(w = mg = 550\;N)$

Normal Reaction $(R) = 450\;N$

Acceleration $(a) = \;?$

We have,

Here, the weight of the body is greater than the normal reaction, hence the elevator is moving downward with acceleration $(a)$. Then for downward motion,

$F = mg - R = 550 - 450 = 100\;N$ 

$m = $$\frac{W}{g} = \frac{550}{10}$$ = 55$

Again,

$F = ma$    ⇒    $a = $$\frac{F}{m}$   =   $\frac{100}{55}$ $= 1.8$

ஃ a = $1.8 m/s^2$ in downward motion.

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A $650\;KW$ power engine of a vehicle of mass $1.5 \; * \; 10^5\; kg$ is rising on an inclined plane of inclination $1$ in $100$ with a constant speed of $60\;km/hr$. Find the frictional force between the wheels of the vehicle and the plane.

Given,

Power of engine $(P) = 650\;kw = 650000\;W$

Mass of vehicle $m = 1.5 \; * \; 10^5\; kg$

Inclination $Sin\;\theta = \frac{1}{100}$

Speed of vehicle $(v) = 60\;km/hr = $$\frac{60\;*\;1000}{60\;*\;60} = \frac{50}{3}m/s$

Frictional force $(f_r) = \;?$

We have, total upwarding force $F = f_r + mg\;sin\;\theta$    ⇒    $\frac{P}{v}$$ = f_r + mg\;sin\;\theta$

or, $f_r = \frac{P}{v} - mg\;sin\;\theta$  =  $\frac{650000\;*\;3}{50}$$ - 1.5\;*\;10^5\;*\;10\;*\;$$\frac{1}{100}$ = $24000\;N$

Hence, fractional force between the wheels of the vehicles and the plane is $24000\;N$.

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A car travelling with a speed of $15\;m/s$ is braked and it slows down with uniform retardation. It covers a distance of $88 \;m$ as its velocity reduces to $7\;m/s$. If the car continues to slow down with the same rate, after what further distance will it be brought to rest?

Given;

Case I:

Initial Velocity $(u) = 15 \; m/s$

Distance Covered $(s) = 88 \; m$

Final Velocity $v = 7 \; m/s$

Retardation $(a) = \; ?$

We have,

$v^2 = u^2 + 2 \; a \; s$    ⇒   $a = $$\frac{v^2 - u^2}{2\;.\;s} = \frac{7^2 - 15^2}{2 \;* \;88}$ $= -1\;m/s^2$

Case II:

Initial Velocity $(u) = 7 \; m/s $

Retardation $(a) = -1 \; m/s^2$

Final Velocity $(v) = 0 \; m/s$

Distance Covered $(s) = \; ?$

Again we have,

$v^2 = u^2 + 2\;a\;s$

$s = $$\frac{v^2 - u^2}{2 * (-1)}$$ = 24.5\;m$

The car brought to be rest at a distance $24.5 \; m$.

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A light rope is attached to a block with mass $4 \; kg$ that rests on a frictionless, horizontal, and surface. The horizontal rope passes over a frictionless pulley and a block with mass $m$ is suspended from the other end. When the blocks are released, the tension in the rope is $10\;N$. Draw free body diagrams and calculate the acceleration of either block and the mass $m$ of the hanging block.

Let $m$ be the mass hanged at the other end of the rope.

Tension in the rope $(T) =10\;N$

Now for $4 \; kg$

$F = T = ma$

$a =$$ \frac{T}{m} = \frac{10}{4}$$ = 2.5\;m/s^2$

For mass $m$

$T = mg -ma$

or, $m(g -a) = T $    ⇒   $ m =$$ \frac{T}{(g - a)} = \frac{10}{10-2.5}$$ = 1.33 \; kg$

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