Explanation
For $$(A):$$ Displacement $$< 0$$ and $$F<0$$ so $$W>0$$
For $$(B):$$ Displacement $$>0$$ and $$F>0$$ so $$W>0$$
For $$(C):$$ Displacement $$>0$$ and $$F<0$$ so $$W<0$$
For $$(D):$$ $$W=\int _{ -2 }^{ 2 }{ -6{ x }^{ 3 } } dx=\left( -\cfrac { 6 }{ 4 } \right) { \left( { x }^{ 4 } \right) }_{ -2 }^{ 2 }=0$$
Given:
$$m_{A} = 20 kg$$
$$m_{B} = 5 kg$$
Kinetic Energy of both bodies A and B are the same.
Thus,
$$\dfrac{1}{2} m_{A} V_{A}^{2} = \dfrac{1}{2} m_{B} V_{B}^{2} \\$$
$$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{m_{B}}{m_{A}} \\$$
$$\dfrac{V_{A}^{2}}{V_{B}^{2}} = \dfrac{5}{20}\\$$
$$\dfrac{V_{A}}{V_{B}} = \dfrac{1}{2}$$
So, option A is correct.
The equation of motion is given as,
$${v^2} = {u^2} + 2as$$
$${\left( 5 \right)^2} = {\left( {20} \right)^2} + 2 \times a \times 100$$
$$a = - 1.875\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$
The force is given as,
$$F = ma$$
$$F = 20 \times \left( { - 1.875} \right)$$
$$F = - 37.5\;{\rm{N}}$$
Thus, the force on the body is $$ - 37.5\;{\rm{N}}$$.
The amount of the work done in raising the glass of the water through a height is calculated by the amount of potential energy
$$W = mgh$$
By substituting the value in the above equation we get
$$W = 0.5 \times 10 \times 0.5$$
Hence the work done in raising the glass is$${\rm{2}}{\rm{.5J}}$$
The initial potential energy of the spring is given as,
$${U_i} = \dfrac{1}{2}k{x^2}$$
$$10 = \dfrac{1}{2}k{\left( {0.3} \right)^2}$$
$$k = \dfrac{{20}}{{0.09}}$$
The final potential energy of the spring is given as,
$${U_f} = \dfrac{1}{2}k{x_1}^2$$
$$ = \dfrac{1}{2} \times \dfrac{{20}}{{0.09}}{\left( {0.45} \right)^2}$$
$$ = 22.5\;{\rm{J}}$$
The amount of work done is given as,
$$W = {U_f} - {U_i}$$
$$ = 22.5 - 10$$
$$ = 12.5\;{\rm{J}}$$
Thus, the amount of work done is $$12.5\;{\rm{J}}$$.
A ball of mass 10 g is projected with an initial velocity of 10 m/s, comes back withd a velocity of 5 m/s at the point of projection. Find the work done by air resistance.(Neglect buoyancy force due to air)
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