What is the minimum work needed to push a 950 kg car 890 m up along a 8.0 degree incline? Assume the effective coefficient of friction retarding the car is 0.22.
Wc = mg = 950kg * 9.8N/kg 9310N. = Weight of car.
Fc = (9310N,8deg.) = Force of car.
Fp = 9310sin8 = 130N. = Force parallel to ramp.
Fv = 9310cos8 = 9219N = Force perpendicular to ramp = The normaql.
Ff = u*Fv = 0.22 * 9219 = 2028N. = Force of friction.
Fn = Fap - Fp - Ff = ma = 0. a = 0.
Fap - 130 - 2028 = 0,
Fap = 130 + 2028 = 2158N = Force applied.
W = F*d = 2158 * 890 = 1,920,620J.
Well, pushing a car uphill sounds like quite the workout! But don't worry, I've got a joke for you while we calculate the minimum work needed.
Why did the car go to therapy? Because it had too many uphill battles to fight!
Now, let's get down to business. To calculate the minimum work needed, we first need to find the force required to overcome the frictional force and pull the car uphill. The force of friction is equal to the coefficient of friction multiplied by the normal force.
The normal force can be calculated using the weight of the car, because on an inclined plane, the weight can be resolved into two components - the perpendicular force (normal force) and the parallel force (the force required to overcome friction).
So, the normal force equals the weight of the car multiplied by the cosine of the angle of the incline.
Now, let's calculate the normal force:
Normal force = (Weight of the car) * (cosine of the angle of incline)
= (950 kg * 9.8 m/s^2) * (cosine of 8.0 degrees)
Next, we calculate the force of friction:
Force of friction = (Coefficient of friction) * (Normal force)
The minimum work required can be calculated using the work-energy principle:
Minimum work = (Force of friction) * (distance)
Plug in the values, and you'll get the answer!
But remember, the joke is on the house. The math, though, may require a little more effort.
To calculate the minimum work needed to push the car up the incline, we need to consider the work done against gravity and the work done against friction.
1. Work done against gravity:
The force of gravity acting on the car can be calculated using the formula: F_gravity = m * g * sin(theta), where m is the mass of the car, g is the acceleration due to gravity (approximately 9.8 m/s^2), and theta is the angle of the incline (8.0 degrees).
F_gravity = 950 kg * 9.8 m/s^2 * sin(8.0 degrees)
2. Work done against friction:
The force of friction can be calculated using the formula: F_friction = m * g * mu, where mu is the coefficient of friction (0.22) and g is the acceleration due to gravity.
F_friction = 950 kg * 9.8 m/s^2 * 0.22
3. Net force:
The net force required to push the car up the incline is the sum of the force required to overcome gravity and the force required to overcome friction.
Net force = F_gravity + F_friction
4. Work done against gravity:
The work done against gravity can be calculated using the formula: Work_gravity = F_gravity * distance, where the distance is given as 890 m.
Work_gravity = F_gravity * 890 m
5. Work done against friction:
The work done against friction can be calculated using the formula: Work_friction = F_friction * distance, where the distance is given as 890 m.
Work_friction = F_friction * 890 m
6. Total work:
The total work needed to push the car up the incline is the sum of the work done against gravity and the work done against friction.
Total work = Work_gravity + Work_friction
Now, let's calculate the values using the given information:
F_gravity = 950 kg * 9.8 m/s^2 * sin(8.0 degrees)
F_friction = 950 kg * 9.8 m/s^2 * 0.22
Net force = F_gravity + F_friction
Work_gravity = F_gravity * 890 m
Work_friction = F_friction * 890 m
Total work = Work_gravity + Work_friction
Please note that the calculations involve trigonometric functions, so it is important to use a calculator that supports these functions.
To find the minimum work needed to push the car up the incline, we need to consider the forces acting on the car and calculate the work done against those forces.
First, let's determine the force of gravity acting on the car. The force of gravity can be calculated using the formula:
Fg = m * g
where:
Fg is the force of gravity
m is the mass of the car (950 kg)
g is the acceleration due to gravity (approximated as 9.8 m/s²)
Fg = 950 kg * 9.8 m/s²
Fg = 9310 N
Next, we need to calculate the component of the force of gravity acting along the incline. This can be done by multiplying the force of gravity by the sine of the angle of the incline:
F_parallel = Fg * sin(θ)
where:
F_parallel is the component of the force of gravity acting along the incline
θ is the angle of the incline (8.0 degrees)
F_parallel = 9310 N * sin(8.0°)
F_parallel = 1288.65 N
Now, let's calculate the frictional force acting against the car. The frictional force can be calculated using the formula:
F_friction = μ * F_norm
where:
F_friction is the frictional force
μ is the coefficient of friction (0.22)
F_norm is the normal force
The normal force can be determined using the formula:
F_norm = Fg * cos(θ)
F_norm = 9310 N * cos(8.0°)
F_norm = 9263.81 N
Finally, we can calculate the frictional force:
F_friction = 0.22 * 9263.81 N
F_friction = 2038 N
The minimum work needed to push the car up the incline can be calculated by adding the work done against gravity to the work done against friction:
Work = Work_gravity + Work_friction
Work_gravity = F_parallel * d
Work_gravity = 1288.65 N * 890 m
Work_gravity = 1145819 J (Joules)
Work_friction = F_friction * d
Work_friction = 2038 N * 890 m
Work_friction = 1813220 J (Joules)
Work = 1145819 J + 1813220 J
Work = 2959039 J (Joules)
Therefore, the minimum work needed to push the 950 kg car 890 m up along the 8.0 degree incline, assuming a coefficient of friction of 0.22, is 2959039 Joules.