Explanation
The average speed is given as,
$${v_{avg}} = \dfrac{s}{t}$$
$${v_{avg}} = \dfrac{{\dfrac{{{V^2}}}{{2\alpha }} + \dfrac{{{V^2}}}{{2\beta }}}}{{\dfrac{V}{\alpha } + \dfrac{V}{\beta }}}$$
$${v_{avg}} = \dfrac{V}{2}$$
When a body moves upward the acceleration due to gravity is,
$${g'} = g + a$$
Hence, the pressure is
$$P = h_0\rho \left( {g + a} \right)$$
$$ = \dfrac{{h_0\left( {g + a} \right)}}{g}$$
Hence, the barometer reading will be more than the measured value at rest.
The momentum of each bullet is given as,
$${P_b} = {m_b}{v_b}$$
$$ = 10 \times {10^{ - 3}} \times 500$$
$$ = 5\;{\rm{kgm/s}}$$
The force on the vehicle will be equal to change in momentum in $$1\;{\rm{sec}}$$.
The force is given as,
$$F = n{P_b}$$
$$ = 10 \times 5$$
$$ = 50\;{\rm{N}}$$
The acceleration of the vehicle is given as,
$$a = \dfrac{F}{{{m_v}}}$$
$$ = \dfrac{{50}}{{2000}}$$
$$ = 0.025\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$
Thus, the acceleration produced in the vehicle is $$0.025\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$.
When the system is released from the position of rest, the equations of motions are given as,
$$4a = 4g - T$$ (1)
$$T = 3a$$ (2)
From the equation (1) and (2) it can be written as,
$$7a = 4g$$
$$a = \dfrac{{4g}}{7}$$
$$a \approx \dfrac{g}{2}$$
Thus, the acceleration of block B is $$\dfrac{g}{2}$$ towards the left side.
Let,
Constant force $$=K_1$$
Variable force $$=-K_2V$$
Net Force $$F=K_1-K_2V$$
Acceleration $$a=\dfrac{K_1-K_2V}{m}$$
As the velocity increases the net force will decrease. When both retarding and constant forces become equal then net force is zero. Hence the velocity is maximum and acceleration will increase from zero to constant.
The acceleration is given as
$$v = u + at$$
$$a = 2\;m/{s^2}$$
The force applied on the horizontal surface is such that it just overcome the static friction
$$F - \mu N = ma$$
$$200 - \mu \cdot 30 \cdot 10 = 30 \times 2$$
The static friction is$$0.47$$
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: I tonne 1000 kg.)
Displacement in first, $$10\,\sec $$
$${{s}_{1}}=\dfrac{1}{2}{{a}_{1}}{{t}_{1}}^{2}=\dfrac{1}{2}\times 2\times {{10}^{2}}=100\,m$$
Speed achieve in first $$10\,\sec $$
$${{v}_{1}}={{u}_{1}}+{{a}_{1}}{{t}_{1}}=0+2\times 10=20\,m/s$$
Displacement in $$30\,\sec $$, acceleration is zero.
$${{s}_{2}}={{v}_{1}}{{t}_{2}}=20\times 30=600\,m$$
Displacement in last due to deacceleration, $${{a}_{3}}=-4\,m/{{s}^{2}}$$
$$ {{v}^{2}}-{{v}_{1}}^{2}=2as $$
$$ {{s}_{3}}=\dfrac{0-{{20}^{2}}}{2\times (-4)}=50\,m $$
Net displacement
$${{S}_{total}}={{s}_{1}}+{{s}_{2}}+{{s}_{3}}=100+600+50=750\,m$$
Total displacement is $$750\,m$$
The time taken to reach the stone on the floor is given as,
$$s = ut + \frac{1}{2}\left( {a + g} \right){t^2}$$
$$3 = 8 \times t + \frac{1}{2}\left( {2 + 10} \right){t^2}$$
$$3 = 8 \times t + 6{t^2}$$
$$t = 0.3052\;\sec $$
Since $$50\;{\rm{N}}$$ force is required for the car to move with constant velocity.
Additional force applied to move a car of $$500\;{\rm{kg}}$$ with acceleration of $$1\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$$is given by
$$F = ma$$
$$F = 500 \times 1$$
$$F = 500\;{\rm{N}}$$
Thus total force applied
$${F_T} = 500 + 50$$
$$F = 550\;{\rm{N}}$$
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