(D) The work done by a constant force is given by the formula $W = F S \cos \theta$,where $F$ is the magnitude of the force,$S$ is the displacement,an

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(D) The work done by a constant force is given by the formula $W = F S \cos \theta$,where $F$ is the magnitude of the force,$S$ is the displacement,and $\theta$ is the angle between the force vector and the displacement vector.Given:Force $F = 10\, N$Displacement $S = 4\, m$Angle $\theta = 60^\circ$Substituting these values into the formula:$W = 10 \times 4 \times \cos 60^\circ$Since $\cos 60^\circ = 0.5$,we have:$W = 10 \times 4 \times 0.5 = 20\, J$.Therefore,the work done by the force is $20\, J$.

Class 11 Physics Work, Energy, Power and Collision Work Done by Constant Force

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$A$ body of mass $5\, kg$ is placed at the origin and can move only on the $x$-axis. $A$ force of $10\, N$ is acting on it in a direction making an angle of $60^\circ$ with the $x$-axis and displaces it along the $x$-axis by $4\, m$. The work done by the force is $...\, J$.

A
$2.5$
B
$7.25$
C
$40$
D
$20$

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Work, Energy, Power and Collision

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Work Done by Constant Force

$A$ block of mass $m$ is kept in an elevator which starts moving downward with an acceleration $a$ as shown in the figure. The block is observed by two observers $A$ and $B$ for a time interval $t_0$. The observer $B$ (who is on the ground) finds that the work done by gravity on the block is:

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$A$ force $\vec F = 3\hat i + c\hat j + 2\hat k$ acting on a particle causes a displacement $\vec S = - 4\hat i + 2\hat j - 3\hat k$. If the work done is $6 \ J$,then the value of $c$ will be:

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$A$ body of mass $m$ moves in a circle of radius $r$ with a constant speed $V$. $A$ force of $mV^2/r$ acts towards the center of the circle. Find the work done by this force when the body covers half the circumference of the circle.

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$A$ simple pendulum bob of mass $m$ is raised to a maximum height $h$. Find the work done by the gravitational force as the pendulum moves from one extreme end to the other.

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$A$ force acts on a $30 \,g$ particle in such a way that the position of the particle as a function of time is given by $x = \alpha t^2$,where $x$ is in metre,$t$ is in seconds and $\alpha = 1 \,m/s^2$. The work done during the first $4 \,s$ is (in $J$)

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$A$ body of mass $5\, kg$ is placed at the origin and can move only on the $x$-axis. $A$ force of $10\, N$ is acting on it in a direction making an angle of $60^\circ$ with the $x$-axis and displaces it along the $x$-axis by $4\, m$. The work done by the force is $...\, J$.

Similar Questions

$A$ block of mass $m$ is kept in an elevator which starts moving downward with an acceleration $a$ as shown in the figure. The block is observed by two observers $A$ and $B$ for a time interval $t_0$. The observer $B$ (who is on the ground) finds that the work done by gravity on the block is:

$A$ force $\vec F = 3\hat i + c\hat j + 2\hat k$ acting on a particle causes a displacement $\vec S = - 4\hat i + 2\hat j - 3\hat k$. If the work done is $6 \ J$,then the value of $c$ will be:

$A$ body of mass $m$ moves in a circle of radius $r$ with a constant speed $V$. $A$ force of $mV^2/r$ acts towards the center of the circle. Find the work done by this force when the body covers half the circumference of the circle.

$A$ simple pendulum bob of mass $m$ is raised to a maximum height $h$. Find the work done by the gravitational force as the pendulum moves from one extreme end to the other.

$A$ force acts on a $30 \,g$ particle in such a way that the position of the particle as a function of time is given by $x = \alpha t^2$,where $x$ is in metre,$t$ is in seconds and $\alpha = 1 \,m/s^2$. The work done during the first $4 \,s$ is (in $J$)

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