Physics Vidyapith ✍️

Formulas and Solutions to Numerical Problems of Power Important Formula of Power: 1.) $P=\frac{Work(W)}{Time(t)}$ 2.) $P=\frac{F.s}{t}$ 3.) $P=\frac{F \: s \: cos \theta}{t}$ 4.) $P=\frac{F \: s }{t} \qquad \left( \because \theta =0^{\circ} \right)$ 5.) $P=F \: v \qquad \left( \because v = \frac{s}{t} \right)$ Numerical Problems and Solutions Q.1 A woman pulls a bucket of water of mass $5 Kg$ from a well which is $10 m$ deep in 10 sec. Calculate the power used by her $(g=10 \: m/sec^{2})$. Solution: Given that: The mass of bucket of water $(m)=5 \: Kg$ Depth of well $(h)=10 \: m$ The time taken to pull a bucket from well $t= 10 \: sec $ The value of gravitational acceleration $g=10 \: m/sec^{2}$ The power used $(P)=?$ Now the power used by her: $P=\frac{W}{T}$ $P=\frac{mgh}{t} \qquad \left( \because W=mgh \right)$ Now Substitute the given values in the equation of power: $P=\frac{5 \times 10 \tim

Solutions to Numerical Problems of the Centripetal Force and Centripetal Acceleration Numerical Problems and Solutions Q.1 An object of mass $0.20 \: Kg$ is being revolved in a circular path of diameter $2.0 \: m$ on a frictionless horizontal plane by means of a string. It performs $10$ revolutions in $3.14 \: s$. Find the centripetal force acting on an object. Solution: Given that: Mass of body $(m)=0.20 \: Kg$ Diameter of Circular path $(d)=2.0: m$ Number of revolutions $(n)=10$ Time taken to complete $(10)$ revolution$(t)=3.14 s$ The centripetal force acting on a body $(F)=?$ Now the centripetal force: $F=m r \omega^2$ $F=m r \left( \frac{2 \pi n}{t} \right)^2$ Now Substitute the given values in the centripetal force formula: $F=0.20 \times 1 \left( \frac{2 \times 3.14 \times 10}{3.14} \right)^2$ $F=0.8 N$ Check Solution Q.2 A car of mass $1200 Kg$ is moving with a speed of $10.5 ms^{-1}$ on a