Area bounded by force time graph with time axis represents impulse. Area of triangle in the figure is ⇒Impulse = ½×2 ×1800 = 1800 kg m/sec

Net force acting on the object is 10 N.

so, acceleration a = F/m = 10 N/10 kg= 1 m/s²

Horizontal component of force is F cosθ = 20√2 cos 45° = 20√2 × 1/√2 = 20 N.

So, Horizontal acceleration =  20N / 10kg  = 2m/s²

According to Newtons second law of motion, Force is time rate of change in momentum.

So, F = change in momentum / time

F= m (vf-vi)/t= 1 kg (0-6)/0.1= -60 N

Given force is F = (3ˆi + 4ˆj +12kˆ ) N , magnitude of force is

⇒F²= (3²+4²+12²) so, F = 13 N

From Newton's second law of motion F= m.a

⇒ 13 N = m.1 so, m=13 kg

Force acting on the object is F = v.(dm/dt)= 5×4 = 20 N

so Acceleration is a= F/m = 20/5 = 4 m/s²

Resultant force on the object will be F= (8² + 6²)½= 10 N

So, acceleration is a = F/m = 10/1 = 10 m/s²

S_n = u + a/2(2n-1) = a/2(2n-1)

S_n+1 = u + a/2(2(n+1)-1) = a/2(2n+1)

S_n / S_n+1 = (2n-1)/(2n+1)

The force required to keep the conveyor belt moving at a constant velocity is given by the rate of change of momentum.

Momentum of sand per second:

$$\text{Force} = \frac{\text{dP}}{\text{dt}} = \frac{\text{dm}}{\text{dt}} \times v$$

Given $\frac{dm}{dt} = 2$ kg/s and $v = 4$ m/s,

$$F = 2 \times 4 = 8 \text{ N}$$

Thus, the additional force required is 8 N.

Solution:

Impulse is given by the change in momentum:

$$J = \Delta p = m \Delta v$$

From the graph, we find velocity before and after $t = 2s$:

• Before $t = 2s$:

  $$v_{\text{initial}} = \frac{x_2 - x_1}{t_2 - t_1} = \frac{0 - 20}{2 - 0} = -10 \text{ m/s}$$

• After $t = 2s$:

  $$v_{\text{final}} = \frac{x_3 - x_2}{t_3 - t_2} = \frac{20 - 0}{4 - 2} = 10 \text{ m/s}$$

Now, impulse:

$$J=m(v_{\text{final}}-v_{\text{initial}})$$

$$J = m (v_{\text{final}} - v_{\text{initial}})$$ $$J = 2 \times (10 - (-10)) = 2 \times 20 = 40 \text{ Ns}$$

Final Answer:

$$\mathbf{40 \text{ Ns}}$$