A car can decelerate at -2.47-m/s2 without skidding when coming to rest on a level road. What would its deceleration (in m/s2) be if the road is inclined at 8.7o and the car moves uphill? (Assume the same static friction coefficient.
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-2.47 - 9.81 sin 8.7
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To solve this problem, we need to break down the gravitational force into two components: one parallel to the road surface (which affects the car's motion) and one perpendicular to the road surface (which is supported by the normal force).
First, let's find the gravitational force acting on the car along the inclined road. The gravitational force can be calculated using the equation:
Fg = m * g
Where Fg is the gravitational force, m is the mass of the car, and g is the acceleration due to gravity (approximately 9.8 m/s^2). Since we are only interested in the component of the gravitational force parallel to the road, we need to multiply the total gravitational force by the sine of the angle of the incline.
F_parallel = Fg * sin(angle)
Next, we need to determine the deceleration of the car along the inclined road when the brakes are applied. The net force acting on the car in the direction opposite to its motion is given by:
Net force = mass * deceleration
In this case, the net force is the difference between the frictional force and the component of the gravitational force parallel to the road. The frictional force can be calculated using the equation:
F_friction = coefficient * normal force
Where F_friction is the frictional force, the coefficient is the static friction coefficient (assumed to be the same), and the normal force can be calculated using the equation:
Normal force = m * g * cos(angle)
Since the car comes to rest, the net force acting on the car is equal to zero:
Net force = F_friction - F_parallel = 0
Therefore, we can calculate the deceleration as follows:
deceleration = F_parallel / m
Substituting the equations above, we have:
deceleration = (Fg * sin(angle)) / m
Now we can solve for the deceleration when the road is inclined at 8.7o and the car moves uphill.