What is the minimum work needed to push a 1050 kg car 460 m up along a 17.5° incline?
(a) Ignore friction
(b) Assume the effective coefficient of friction retarding the car is 0.30
work= weight*height= m*g* 460SinTheta.
Now with friction, add to the basic work in a), the work done on friction, which is normal component of weight*coefficent*distance
or
friction= mg*CosTheta*mu*460
I took m*g*460 sintheta
I used 1050*9.8* 460sin 17.5 and that's not the right answer I think I messed up somewhere.
I get for a) 1.050E3*9.8*4.6E2*sin17.5=6.97E5 joules
okay i got 14.2e5 for a which is right but for b I keep getting it wrong, i used mg*CosTheta*mu*460
and I used 1.050e3*9.8*cos17.5*.30*460
but that aswer is not right I am not sure is the friction .30 =mu?
To calculate the minimum work needed to push a car up an incline, we need to consider the forces acting on the car.
(a) When ignoring friction, the only force acting on the car is its weight. The work done against gravity can be calculated using the formula:
Work = Force x Distance x cos(theta)
In this case, the force is equal to the weight of the car (mg), the distance is the displacement along the incline, and theta is the angle of the incline.
Given:
Mass of the car (m) = 1050 kg
Distance (d) = 460 m
Angle of the incline (theta) = 17.5°
We can calculate the work as follows:
Weight of the car = mg
Weight = 1050 kg x 9.8 m/s^2 (acceleration due to gravity)
Weight ≈ 10290 N
Work = 10290 N x 460 m x cos(17.5°)
Therefore, the minimum work needed to push the car up the incline when ignoring friction is given by the above calculation.
(b) In the case where friction is considered, we need to account for the work done against friction as well. The work done against friction is given by:
Work_friction = Frictional force x Distance
The frictional force can be calculated using the formula:
Frictional force = coefficient of friction x Normal force
The normal force is the force exerted by the incline on the car and is equal to mg cos(theta).
Given:
Mass of the car (m) = 1050 kg
Distance (d) = 460 m
Angle of the incline (theta) = 17.5°
Coefficient of friction (μ) = 0.30
We can calculate the work against friction as follows:
Normal force = mg cos(theta)
Normal force = 1050 kg x 9.8 m/s^2 x cos(17.5°)
Frictional force = μ x Normal force
Frictional force = 0.30 x (1050 kg x 9.8 m/s^2 x cos(17.5°))
Work_friction = Frictional force x Distance
Work_friction = (0.30 x (1050 kg x 9.8 m/s^2 x cos(17.5°))) x 460 m
Therefore, the minimum work needed to push the car up the incline, considering the friction, is given by the above calculation.