Question

In a tractor-pull competition, a tactor applies a force of 1.3 kN to the sled, which has a mass of 1.1x10^4 kg. At that point, the coefficent of friction between the sled and the ground is 0.80. What is the acceleration of the sled?

Answer 1

In a tractor-pull competition, a tractor applies a force of 1.3KN to the sled, which has mass 1.1* 10^4 Kg. At that point, the co-efficient of kinetic friction between the sled and the ground has increased to .80. What is the acceleration of the sled? Explain the significance of the sign of the acceleration.

I got -7.73 m/s2 but my answer is way off...

Answer 2

ma = F - mu m g, where F is the force applied by truck on the
sledge and mu, coefficient of friction, or
a = F/m - mu g= 1.3x10^3/1.1x10^4 - 0.80x9.8
= - 7.71 m/s^2

Answer 3

this belongs in the physics forum

Answer 4

Thanks for your quick answer but you are off. A = 0.61 m/s^2 in the back of the book. The physics forum doesn't exist. =/

Answer 5

Nevermind, found it.