A 1000kg car is travelling down at a 25° slope. At the instant that time speed is 13m/s,the driver applies the brakes,what constant forces parallel to the road,must be provided by the brakes if the car is to stop in 90 meters
To find the constant force required to stop the car in 90 meters, we need to break down the forces acting on the car.
First, we need to find the gravitational force acting on the car along the slope. The gravitational force Fg can be calculated as:
Fg = m*g*sinθ
Where:
m = mass of the car = 1000 kg
g = acceleration due to gravity = 9.81 m/s²
θ = angle of the slope = 25°
Plugging in the values, we get:
Fg = 1000*9.81*sin(25°)
≈ 4257.68 N
Next, we need to calculate the frictional force (force provided by the brakes) required to stop the car. The frictional force Ff can be calculated using the equation:
Ff = m*a
Where:
a = acceleration of the car
Vf = final velocity = 0 m/s
To find the acceleration, we can use the kinematic equation:
Vf² = Vi² + 2*a*s
Where:
Vi = initial velocity = 13 m/s
s = stopping distance = 90 m
Rearranging the equation to solve for acceleration a, we get:
a = (Vf² - Vi²) / 2*s
= (0 - 13²) / (2*90)
= -169 / 180
≈ -0.94 m/s²
Now, plugging in the acceleration value along with the mass, we can find the frictional force required to stop the car:
Ff = 1000 * (-0.94)
≈ -940 N
Since frictional force always opposes motion, the required force provided by the brakes would be the negative of this value. Therefore, the constant force parallel to the road that the brakes must provide to stop the car in 90 meters is approximately 940 N.