Amotorvehicle of mass 1200kg is travelling along alevel road at 54km/hr.The breaks are then applied and the speed of the vehicle reduced to 18km/hr over adistance of 50M .if the afriction motion resistant is constant at 160N .calculate the effective braking force of the brakes. SLN.Mass =1200kg vel=54km/hr*1000 divide by 3600=15m/s speed of vehicle=5m/s distance=50m 10*1600*50=800000joules

How you come up whith this answer?

To calculate the effective braking force, we need to first calculate the work done by the friction force and then use that to determine the braking force. Here's how:

1. Calculate the initial kinetic energy (KE₁) of the vehicle using the mass (m) and initial speed (v₁):
KE₁ = (1/2) * m * v₁²
KE₁ = (1/2) * 1200 kg * (15 m/s)²

2. Calculate the final kinetic energy (KE₂) of the vehicle using the mass (m) and final speed (v₂):
KE₂ = (1/2) * m * v₂²
KE₂ = (1/2) * 1200 kg * (5 m/s)²

3. Calculate the work done (W) by the friction force, which is equal to the change in kinetic energy:
W = KE₁ - KE₂

4. Use the work done (W) and the distance (d) to calculate the effective braking force (F):
W = F * d

Now let's calculate each step:

1. Calculate the initial kinetic energy (KE₁):
KE₁ = (1/2) * 1200 kg * (15 m/s)² (Note: Make sure to square the speed in m/s)
KE₁ = 1/2 * 1200 kg * 225 m²/s²
KE₁ = 135000 J

2. Calculate the final kinetic energy (KE₂):
KE₂ = (1/2) * 1200 kg * (5 m/s)²
KE₂ = 1/2 * 1200 kg * 25 m²/s²
KE₂ = 15000 J

3. Calculate the work done (W) by the friction force:
W = KE₁ - KE₂
W = 135000 J - 15000 J
W = 120000 J

4. Calculate the effective braking force (F):
W = F * d
120000 J = F * 50 m
F = 120000 J / 50 m
F = 2400 N

Therefore, the effective braking force of the brakes is 2400 Newtons.