A machine have velocity ratio 5 requires a force of 600joule to rise a load of 400 newton.if the load moves a distance of 0.5 metres, what is mechanical advantages and efficiency?

First, let's calculate the work done by the machine:

Work = Force x Distance
Work = 400 N x 0.5 m
Work = 200 Joules

Now, let's calculate the mechanical advantage:

Mechanical Advantage = Output Force / Input Force
Mechanical Advantage = 400 N / 600 N
Mechanical Advantage = 0.67

Next, let's calculate the efficiency:

Efficiency = (Output Work / Input Work) x 100%
Efficiency = (200 J / 600 J) x 100%
Efficiency = 33.33%

Therefore, the mechanical advantage is 0.67 and the efficiency is 33.33%.

Given:

- Velocity ratio (VR) = 5
- Force applied (F) = 600 N
- Load (W) = 400 N
- Distance moved by the load (d) = 0.5 m

To find:

- Mechanical advantage (MA)
- Efficiency (η)

Mechanical Advantage (MA):

MA is the ratio of the load lifted to the force applied.

MA = W/F
MA = 400 N / 600 N
MA = 0.67

Efficiency (η):

Efficiency is the ratio of output work (work done by the load) to input work (work done by the effort).

Efficiency (η) = (Load x Distance moved by the load) / (Force applied x Distance moved by the effort)

η = (400 N x 0.5 m) / (600 N x VR x 0.5 m)
η = 400 Nm / (600 Nm x 5)
η = 0.14

Therefore, the mechanical advantage of the machine is 0.67, and its efficiency is 14%.