A car can decelerate at -3.40 m/s2 without skidding when coming to rest on a level road. What would be the magnitude of its deceleration if the road were inclined at 14° uphill? Assume the same static friction coefficient.
To find the magnitude of the car's deceleration if the road were inclined at 14° uphill, we need to find the component of gravitational force that acts downhill.
The gravitational force acting downhill can be calculated using the formula:
F_downhill = m * g * sin(θ)
Where:
m = mass of the car
g = acceleration due to gravity (approximately 9.8 m/s^2)
θ = angle of the road (14°)
Since the static friction coefficient does not change, the maximum static friction force can be calculated using the formula:
F_friction = μ * m * g
where μ is the static friction coefficient.
Since the car is decelerating without skidding, the maximum static friction force must equal the gravitational force acting downhill:
F_friction = F_downhill
Therefore, we can set up the following equation:
μ * m * g = m * g * sin(θ)
Canceling out the mass and acceleration due to gravity, we can solve for the static friction coefficient:
μ = sin(θ)
So, the magnitude of the car's deceleration would be equal to the magnitude of gravitational acceleration acting downhill:
a = g * sin(θ)
Plugging in the values:
a = 9.8 m/s^2 * sin(14°)
Calculating this, we find:
a ≈ 2.58 m/s^2
Therefore, the magnitude of the car's deceleration would be approximately 2.58 m/s^2 if the road were inclined at 14° uphill.
To find the magnitude of the deceleration of the car on an inclined road, we need to consider the forces acting on the car.
The main force affecting the car's motion on the inclined road is the force of static friction. The force of static friction acts in the opposite direction of the car's motion and prevents it from sliding or skidding on the road.
In this case, we assume that the static friction coefficient between the car's tires and the road remains the same. The static friction force can be calculated using the formula:
fs = μs * N
Where fs is the static friction force, μs is the static friction coefficient, and N is the normal force.
The normal force is the perpendicular force exerted by the road on the car. On a level road, the normal force is equal to the car's weight, which can be calculated as:
N = m * g
Where m is the mass of the car and g is the acceleration due to gravity.
On an inclined road, the normal force is reduced due to the component of the car's weight acting parallel to the slope. The normal force can be calculated as:
N = m * g * cos(θ)
Where θ is the angle of inclination, which in this case is 14°.
Now we can find the static friction force on the inclined road:
fs = μs * m * g * cos(θ)
The deceleration of the car is related to the static friction force by Newton's second law:
fs = m * a
Where a is the deceleration. Rearranging this equation, we can solve for the deceleration:
a = fs / m
Substituting the expression for fs, we get:
a = (μs * m * g * cos(θ)) / m
The mass of the car cancels out, and the final expression for the deceleration is:
a = μs * g * cos(θ)
Now, plug in the given values:
μs = static friction coefficient (assume it remains the same)
g = acceleration due to gravity (approximately 9.8 m/s²)
θ = 14°
a = μs * g * cos(θ)
Note that the negative deceleration mentioned in the question does not affect the magnitude of the deceleration, so we consider the absolute value.
Calculating the equation, we get:
a = |(-3.40) * 9.8 * cos(14°)|
So the magnitude of the deceleration of the car on the inclined road would be approximately 35.87 m/s².
well, the braking force is F=ma=-3.40m
and that is equal to mu*mg
so mu=3.40/9,8
Now going on a hill, the normal force is less (normal force=mg*Cos14), so braking force is mu*normal= 3.40/9.8*mgCos14=3.40*mCos14
That is the breaking force.
However, on a hill, gravity also adds a force down the hill mgSin14
This force aids braking (going up hill), or resists braking.
Going downhill:
3.40*m*Cos14-mgSin14=ma
a= 3.40Cos14-9.8Sin14
Going uphill:
3.40*m*Cos14+9.8sin14=ma
a= 3.40Cos14+9.8Sin14