84. two blocks, one of mass 5.0 kg and the other of
mass 3.0 kg, are tied together with a massless rope
as in figure 28. this rope is strung over a massless,
resistance-free pulley. the blocks are released from
rest. find the following:
a. the tension in the rope
b. the acceleration of the blocks
hint: you will need to solve two simultaneous equations.

a) 37 N

b) 2.5 m/s²

Explanation:

Let's say m₁ is the 5 kg mass, and m₂ is the 3 kg mass.

Draw a free body diagram for each block.  For both, there are two forces: weight down and tension up.

Apply Newton's second law to the 5 kg mass in the y direction (remember it is accelerating downward).

∑F = ma

T − m₁g = m₁(-a)

m₁g − T = m₁a

Apply Newton's second law to the 3 kg mass in the y direction:

∑F = ma

T − m₂g = m₂a

a) Solve the system of equations.  First, solve for a in each equation, then set equal and solve for T.

a = (m₁g − T) / m₁

a = (T − m₂g) / m₂

(m₁g − T) / m₁ = (T − m₂g) / m₂

m₂ (m₁g − T) = m₁ (T − m₂g)

m₁m₂g − Tm₂ = m₁T − m₁m₂g

2m₁m₂g = (m₁ + m₂)T

T = 2m₁m₂g / (m₁ + m₂)

Given m₁ = 5 kg and m₂ = 3 kg:

T = 2 (5 kg) (3 kg) (9.8 m/s²) / (5 kg + 3 kg)

T = 36.75 N

Rounded to two significant figures, the tension is 37 N.

b) Solve the system of equations.  First, solve for T in each equation, then set equal and solve for a.

T = m₁g − m₁a

T = m₂g + m₂a

m₁g − m₁a = m₂g + m₂a

m₁g − m₂g = m₁a + m₂a

(m₁ − m₂)g = (m₁ + m₂)a

a = (m₁ − m₂)g / (m₁ + m₂)

Given m₁ = 5 kg and m₂ = 3 kg:

a = (5 kg − 3 kg) (9.8 m/s²) / (5 kg + 3 kg)

a = 2.45 m/s²

Rounded to two significant figures, the magnitude of the acceleration of the blocks is 2.5 m/s².

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min

answeralert drivers who drift off the road;