Question

Two blocks of weight 3.0N and 7.0N are connected by a massless string and slide down a 30 degree inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.13; that between the heavier block and the plane is 0.31. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the string

Answer 1

(a) 2.793 m/s^2

(b) 0.335 N

Step-by-step explanation:

Let a be the acceleration in the system and T be the tension in the string.

W1 = 3 N

W2 = 7 N

m1 = 3 /1 0 = 0.3 kg

m2 = 7 / 10 = 0.7 kg

θ = 30°

μ1 = 0.13, μ2 = 0.31

Let f1 be the friction force acting on block 1 and f2 be the friction force acting on block 2.

By the laws of friction

f1 = μ1 x N1

Where, N1 be the normal reaction acting on block 1.

So, f1 = 0.13 x W1 Cosθ = 0.13 x 3 x cos 30 = 0.337 N

By the laws of friction

f2 = μ2 x N2

Where, N2 be the normal reaction acting on block 2.

So, f2 = 0.31 x W2 Cosθ = 0.31 x 7 x cos 30 = 1.88 N

Apply Newton's second law for both the blocks

W1 Sin30 - T - f1 = m1 a

3 Sin 30 - T - 0.337 = 0.3 x a

1.163 - T = 0.3 a ..... (1)

W2 Sin30 + T - f2 = m2 a

7 Sin30 + T - 1.88 = 0.7 x a

1.62 + T = 0.7 a ..... (2)

By solving equation (1) and (2) we get

a = 2.793 m/s^2

(b) Put the value of a in equation (2), we get

1.62 + T = 0.7 x 2.793

T = 0.335 N

Answer 2

a. a =
`6.41 m/s^2`

b. T = -0.81 N

Step-by-step explanation:

Given,

• weight of the lighter block =
  `w_1\ =\ 3.0\ N`
• weight of the heavier block =
  `w_2\ =\ 7.0\ N`
• inclination angle =
  `\theta\ =\ 30^o`
• coefficient of kinetic friction between the lighter block and the surface =
  `\mu_1\ =\ 0.13`

• coefficient of kinetic friction between the heavier block and the surface =
  `\mu_2\ =\ 0.31`
• friction force on the lighter block =
  `f_1\ =\ \mu_1N_1\ =\ \mu_1 w_1cos\theta`
• friction force on the heavier block =
  `f_2\ =\ \mu_2N_2\ =\ \mu_2w_2cos\theta`

Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.

From the f.b.d. of the lighter block,

`w_1sin\theta\ -\ T\ -\ f_1\ =\ (w_1a)/(g)\\\Rightarrow T\ =\ w_1sin\theta\ -\ (w_1a)/(g)\ -\ f_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)`

From the f.b.d. of the heavier block,

`w_2sin\theta\ +\ T\ -\ f_2\ =\ (w_2a)/(g)\\\Rightarrow T\ =\ (w_2a)/(g)\ -\ w_2sin\theta\ -\ f_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (2)`

From eqn (1) and (2), we get,

`w_1sin\theta\ -\ (w_1a)/(g)\ -\ f_1\ =\ (w_2a)/(g)\ -\ w_2sin\theta\ -\ f_2\\\Rightarrow w_1gsin\theta \ -\ w_1a\ -\ \mu_1w_1gcos\theta\ =\ w_2a\ -\ w_2gsin\theta\ -\ \mu_2 w_2gcos\theta\\`

`\Rightarrow a\ =\ (g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta))/(w_1\ +\ w_2)\\\Rightarrow a\ =\ (g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)])/(w_1\ +\ w_2)\\`

`\Rightarrow a\ =\ (9.81* [sin30^o* (3.0\ +\ 7.0)\ +\ cos30^o* (0.31* 7.0\ -\ 0.13* 3.0)])/(3.0\ +\ 7.0)\\\Rightarrow a\ =\ 6.4\ m/s.`

part (b)

From the eqn (2), we get,
`T\ =\ (w_2a)/(g)\ -\ w_2sin\theta\ -\ \mu_2w_2cos\theta\\\Rightarrow T\ =\ (7.0* 6.41)/(9.81)\ -\ 7.0* sin30^o\ -\ 0.31* 7.0* cos30^o\\\Rightarrow T\ =\ -0.81\ N`