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?
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To find the force that moves the truck down the hill, we can use the force of gravity acting on the truck. The force of gravity is given by the weight of the truck, which can be calculated using the formula:
Weight = mass * acceleration due to gravity
The acceleration due to gravity is approximately 9.8 m/s^2.
a) Finding the force that moves the truck down the hill:
Weight = mass * acceleration due to gravity
Weight = 1500 kg * 9.8 m/s^2
Weight = 14,700 N
Therefore, the force that moves the truck down the hill, assuming friction is negligible, is 14,700 Newtons.
b) To find the speed of the truck after 50 seconds, we can use the formula for velocity:
Velocity = initial velocity + (acceleration * time)
Since the truck starts from rest, the initial velocity is 0 m/s. Also, since the slope of the hill is at an angle, we need to find the component of acceleration due to gravity that acts parallel to the slope:
Acceleration = acceleration due to gravity * sin(slope angle)
Using the given slope angle of 7 degrees, the acceleration becomes:
Acceleration = 9.8 m/s^2 * sin(7 degrees)
Acceleration ≈ 9.8 m/s^2 * 0.122
Acceleration ≈ 1.196 m/s^2
Now we can find the velocity after 50 seconds:
Velocity = 0 m/s + (1.196 m/s^2 * 50 s)
Velocity = 0 m/s + 59.8 m/s
Velocity = 59.8 m/s
Therefore, the speed of the truck after 50 seconds is approximately 59.8 meters per second.