PhysicsQ101–102 of 200 questions
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101
PhysicsEasyMCQTS EAMCET · 2018
$A$ small body of mass $500 \,g$ moves on a rough horizontal surface before finally stopping. The initial velocity of the body is $2 \,m/s$ and the coefficient of friction is $0.3$. Find the absolute value of the average power developed by the frictional force during the time of motion. (Take $g = 10 \,m/s^2$) (in $\,W$)
A
$1$
B
$1.5$
C
$2$
D
$2.5$
Solution
(B) Since the retardation is constant,the average velocity is given by:
$v_{av} = \frac{u + v}{2} = \frac{2 + 0}{2} = 1 \,m/s$
The frictional force $f$ is given by:
$f = \mu N = \mu mg$
Given:
Coefficient of friction,$\mu = 0.3$
Mass of body,$m = 500 \,g = 0.5 \,kg$
Acceleration due to gravity,$g = 10 \,m/s^2$
Substituting the values:
$f = 0.3 \times 0.5 \times 10 = 1.5 \,N$
The average power $P_{av}$ developed by the frictional force is:
$P_{av} = f \times v_{av}$
$P_{av} = 1.5 \,N \times 1 \,m/s = 1.5 \,W$
Thus,the absolute value of the average power is $1.5 \,W$.
$v_{av} = \frac{u + v}{2} = \frac{2 + 0}{2} = 1 \,m/s$
The frictional force $f$ is given by:
$f = \mu N = \mu mg$
Given:
Coefficient of friction,$\mu = 0.3$
Mass of body,$m = 500 \,g = 0.5 \,kg$
Acceleration due to gravity,$g = 10 \,m/s^2$
Substituting the values:
$f = 0.3 \times 0.5 \times 10 = 1.5 \,N$
The average power $P_{av}$ developed by the frictional force is:
$P_{av} = f \times v_{av}$
$P_{av} = 1.5 \,N \times 1 \,m/s = 1.5 \,W$
Thus,the absolute value of the average power is $1.5 \,W$.
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102
PhysicsEasyMCQTS EAMCET · 2018
$A$ mechanical system consists of two springs with stiffness coefficients $k_1$ and $k_2$ connected in series. The minimum work to be performed on the system to stretch it by a total displacement $\Delta$ is
A
$\frac{1}{2}\left(\frac{k_1 k_2}{k_1+k_2}\right) \Delta^2$
B
$k_1 k_2 \Delta^2$
C
$\left(\frac{k_1 k_2}{k_1+k_2}\right) \Delta^2$
D
$\left(\frac{k_1 k_2}{k_1+k_2}\right) \Delta$
Solution
(A) When two springs with stiffness constants $k_1$ and $k_2$ are connected in series,the equivalent spring constant $K_{\text{eq}}$ is given by the formula:
$\frac{1}{K_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{k_1 + k_2}{k_1 k_2}$
Therefore,$K_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2}$.
The work done $W$ to stretch a spring system by a total displacement $\Delta$ is equal to the potential energy stored in the equivalent spring:
$W = \frac{1}{2} K_{\text{eq}} \Delta^2$
Substituting the value of $K_{\text{eq}}$:
$W = \frac{1}{2} \left( \frac{k_1 k_2}{k_1 + k_2} \right) \Delta^2$.
$\frac{1}{K_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{k_1 + k_2}{k_1 k_2}$
Therefore,$K_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2}$.
The work done $W$ to stretch a spring system by a total displacement $\Delta$ is equal to the potential energy stored in the equivalent spring:
$W = \frac{1}{2} K_{\text{eq}} \Delta^2$
Substituting the value of $K_{\text{eq}}$:
$W = \frac{1}{2} \left( \frac{k_1 k_2}{k_1 + k_2} \right) \Delta^2$.
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