Solve a problem of your own!
Download the Studdy App!

Math

Math Snap

PROBLEM

(II) What is the minimum work needed to push a 950kg950-\mathrm{kg} car 510 m up along a 9.09.0^{\circ} incline? Ignore friction.

STEP 1

1. The car has a mass of 950kg 950 \, \text{kg} .
2. The incline is 9.0 9.0^\circ .
3. The distance along the incline is 510m 510 \, \text{m} .
4. Friction is ignored.
5. Work is calculated as the force parallel to the incline times the distance.

STEP 2

1. Determine the gravitational force acting on the car.
2. Calculate the component of the gravitational force parallel to the incline.
3. Calculate the work done to move the car up the incline.

STEP 3

Determine the gravitational force acting on the car:
Fgravity=mg F_{\text{gravity}} = m \cdot g where m=950kg m = 950 \, \text{kg} and g=9.8m/s2 g = 9.8 \, \text{m/s}^2 .
Fgravity=950kg×9.8m/s2=9310N F_{\text{gravity}} = 950 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 9310 \, \text{N}

STEP 4

Calculate the component of the gravitational force parallel to the incline:
Fparallel=Fgravitysin(θ) F_{\text{parallel}} = F_{\text{gravity}} \cdot \sin(\theta) where θ=9.0 \theta = 9.0^\circ .
Fparallel=9310Nsin(9.0) F_{\text{parallel}} = 9310 \, \text{N} \cdot \sin(9.0^\circ) Calculate sin(9.0) \sin(9.0^\circ) :
sin(9.0)0.1564 \sin(9.0^\circ) \approx 0.1564 Fparallel=9310N×0.15641456N F_{\text{parallel}} = 9310 \, \text{N} \times 0.1564 \approx 1456 \, \text{N}

SOLUTION

Calculate the work done to move the car up the incline:
W=Fparalleld W = F_{\text{parallel}} \cdot d where d=510m d = 510 \, \text{m} .
W=1456N×510m W = 1456 \, \text{N} \times 510 \, \text{m} W742560J W \approx 742560 \, \text{J} The minimum work needed to push the car up the incline is:
742560J \boxed{742560 \, \text{J}}

Was this helpful?
banner

Start understanding anything

Get started now for free.

OverviewParentsContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord