A3047.8 kg truck has lost its brakes coming down a mountain. fortunately, there is a ramp of thick gravel inclined at 9.5 degrees that the driver can use to stop the truck. the truck hits the ramp moving at 20.68 m/s and then travels 26.6 meters up the ramp before it comes to a stop. what work is done by the gravel to stop the truck?

The work done by the gravel to stop the truck is 520.44 kJ

Explanation:

Step 1: Data given

Mass of the truck = 3047.8 kg

The ramp has an angle of 9.5 °

Velocity of  the truck = 20.68 m/s

distance = 26.6 meters

Step 2: Calculate initial kinetic energy

sin 9.5° = 0.165

h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m

Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ  = initial kinetic energy

Step 3: Calculate potential energy

Epot = U = m*g*h = 3047.8*9.81*4.39 = 131256.25 Joule = 131.26 kJ

Step 4:  What work is done by the truck on the gravel?  

Frictional energy Ef = 651.7 kJ - 131.26 kJ = 520.44 kJ

neither accurate nor precise.

i think it would be 5