a truck with a mass of 1,500 kg is parked on a hill with a slope of 7 degrees. the parking break is released and the truck starts to roll down the hill.
a) assuming friction is negligible what will by the force that moves the car down the hill?
b) what will the speed of the carbs after50 seconds?
friction was neglibible but it was parked on the hill. Go figure.
weight down the hill: mgsin7
that is the force down the hill.
a= F/m
vf=at=F/m * time
a) The force that moves the truck down the hill is the force due to gravity. To calculate this force, we need to find the component of the gravitational force that acts parallel to the slope. This can be done using the formula:
Force = mass * acceleration
The acceleration in this case is the component of the acceleration due to gravity that acts parallel to the slope. The formula for this is:
Acceleration = gravity * sin(angle)
Substituting the given values, we have:
Acceleration = 9.8 m/s² * sin(7°)
Now we can calculate the force:
Force = mass * acceleration
Force = 1,500 kg * (9.8 m/s² * sin(7°))
b) To calculate the speed of the truck after 50 seconds, we need to consider the motion with constant acceleration. In this case, the acceleration is the same as the force calculated in part a) divided by the mass of the truck.
Acceleration = Force / mass
Once we have the acceleration, we can use the following formula to calculate the final speed:
Final speed = Initial speed + (acceleration * time)
Since the truck starts from rest (initial speed = 0), we have:
Final speed = acceleration * time
Substituting the values, we get:
Final speed = (Force / mass) * time
Final speed = (1,500 kg * (9.8 m/s² * sin(7°)) / 1,500 kg) * 50 s
Simplifying the expression:
Final speed = 9.8 m/s² * sin(7°) * 50 s