Question

find acceleration (a) and tension (T)​

Answer 1

Answer:

7 m/s² and 120 N

Explanation:

Let m₁ = 10 kg

m₂ = 40 kg

θ = 30º

Let a be the acceleration, T be the tension

Force balance equation of m₂

m₂g - T = m₂a .......(1)

Force balance equation of m₁

Forces acting on m₁ are

1. Tension T in upward incline direction

2. friction force in downward incline direction, f

3. Normal reaction perpendiculat to plane, N

4. Weight due to mass, m₁g. This force is resolved into m₁gsinθ in downward incline direction and m₁gcosθ perpendiculat to the plane.

Balancing the forces we get

T - m₁gsinθ - f = m₁a .........(2)

N = m₁gcosθ

We know that

f = μΝ = μm₁gcosθ

Equation (2) becomes

T - m₁gsinθ - μm₁gcosθ = m₁a

or T = m₁gsinθ + μm₁gcosθ + m₁a

Calculation of a

Substituting the above value of T in (1), we get

m₂g - m₁gsinθ - μm₁gcosθ - m₁a = m₂a

or m₂g - m₁gsinθ - μm₁gcosθ = (m₂ + m₁)a

or a = (m₂g - m₁gsinθ - μm₁gcosθ)/(m₂ + m₁)

Substituting given values and neglecting friction

a = (40g - 10gsin(30))/50

= (40g - 5g)/50

= 35g/50

= 35*10/50 ( Taking g = 10 m/s² )

= 7 m/s²

Calculation of T

Using above value of a in (1) we get

m₂g - T = m₂(m₂g - m₁gsinθ - μm₁gcosθ)/(m₂ + m₁)

m₂g - m₂(m₂g -m₁gsinθ - μm₁gcosθ)/(m₂ + m₁) = T

T(m₂ + m₁) = m₂g(m₂ + m₁) - m₂(m₂g - m₁gsinθ - μm₁gcosθ

T(m₂ + m₁) = (m₂m₁g + m₂m₁gsinθ + μm₂m₁gcosθ)

T = m₁m₂g(1 + sinθ + μcosθ)/(m₁ + m₂)

Substituting values given and neglecting friction

T = 10*40g(1+sin(30))/50

= 400g*(3/2)/50

= 12g

= 120 N

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Answer 2

Answer:

A car and motorcycle racing from rest along a runway the motorcycle take lead first but then car overtake motorcycle .The motorcycle first take lead because it has acceleration 8.40 m/s²,greater than car acceleration 5.60 m/s² but motorcycle looses to car because motorcycle reaches its greatest speed 58.8 m/s before car reaches its greatest speed 106 m/s