We have an answer to "The rope and pulley have negligible mass, and..."

Carole Yarbrough

Answered question

2021-12-16

The rope and pulley have negligible mass, and the pulley is frictionless. The coefficient of kinetic friction between the
8.00-kg block and the tabletop is $μ k = 0.250$ . The blocks are released from rest. Use energy methods to calculate the speed of the 6.00-kg block after it has descended 1.50 m.

Answer & Explanation

Bertha Jordan

Beginner2021-12-17Added 37 answers

Given:

d = 1.5 m

μ_{k} = 0.25

m_{1} = 8.0 k g

m_{2} = 6.0 k g

Required: v
For horizontal mass
According to work-energy theorem the velocity is given by

W_{\neq t} = \frac{1}{2} m_{1} v^{2} \Rightarrow (1)

But the net work is also given by

W_{\neq t} = W_{t} - W_{f} = W_{t} - μ_{k} m g d \Rightarrow (2)

Where :

W_{t}

is the work due to tension.
From (1) and (2) get that

W_{t} - μ_{k} m_{1} g d = \frac{1}{2} m_{1} v^{2} \Rightarrow (3)

For vertical mass
According to work-energy theorem the velocity is given by

W_{\neq t} = \frac{1}{2} m_{2} v^{2} \Rightarrow (4)

But the net work is also given by

W_{\neq t} = W_{w} - W_{t} = m_{2} g d - W_{t} \Rightarrow (5)

From (4) and (5) get that

- W_{t} + m_{2} g d = \frac{1}{2} m_{2} v^{2} \Rightarrow (6)

Adding (3) and (6) to get

m_{2} g d - μ_{k} m_{1} g d = \frac{1}{2} (m_{1} + m_{2}) v^{2} R i h > \Rightarrow (7)

Substitution in (7) to get that

v^{2} = \frac{6 \times 9.8 \times 1.5 - 8 \times 9.8 \times 1.5 \times 0.25}{7} \Rightarrow v = 2.9 \frac{m}{s}

Mollie Nash

Beginner2021-12-18Added 33 answers

Step1
Given that
mass

m_{1} = 6

kg
mass

m_{2} = 8

kg
coefficient of kinetic friction

μ_{k} = 0.250

displacement of mass 6 kg y=1.50m
Step2
Assume that the uniform acceleration in both masses is "a" and tension in the rope is "T".
Now for mass

m_{1}

effective downward force = 6g - T

m_{1} a = 6 g - T

6 a = 6 g - T

Now for mass

m_{2}

friction force ( towards left)

f_{k} = ∣ μ_{k} m_{2} g

f_{k} = (0.25) (8) (g)

f_{k} = 2 g

effective force on

m_{2} ( towards right) = T - f_{k}

m_{2} a = T - f_{k}

8 a = T - 2 g

Adding equation (1) and (2) as:

(6 a) + (8 a) = (6 g - T) + (T - 2 g)

14 a = 4 g

a = \frac{4 g}{14}

a = \frac{2 g}{7}

Step3
Effective force on

m_{1}

F_{1} = 6 a

F_{1} = (6) (\frac{2 g}{7})

F_{1} = \frac{12 g}{7}

Work done on m, is

W = F_{1} y

W = (\frac{12 g}{7}) (1.5)

W = \frac{18 g}{7}

Step4
Now by the use of work energy theorem
Work done = change in kinetic energy

W = \frac{1}{2} m v^{2}

where v = speed

\frac{18 g}{7} = \frac{1}{2} (6) v^{2}

v^{2} = \frac{36 g}{42}

g= gravitational acceleration (

9.8 \frac{m}{s^{2}}

)

v = \sqrt{\frac{(36) (9.8)}{42}}

v = 2.989 \frac{m}{s}

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