Astuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. the cable is parallel to the ground. the mass of the stuntman is 119 kg, and the coefficient of kinetic friction between the road and him is 0.697. find the tension in the cable.

T = 812.8414 N

Explanation:

Using the law of newton we found the vertical(y) and horizontal(x) forces as:

∑ = T - = ma

Where T is the tension, is the friction force, m is the mass of the stuntman and a is the aceleration of the stuntman.

but a is equal to 0 because he is moving at a constant velocity, so:

T - = 0

T =

Also,

=

where is the coefficient of kinetic friction and N is the normal force.

For find N we use:

∑ N - mg = 0

N = mg

N = (119)(9.8)

N = 1166.2

Finally we solve for T as:

T =

T = (0.697)(1166.2)

T = 812.8414 N

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14