physics help please?

Question

1) a person with a mass of 64 kg sits at a distance of 2.7 m from the pivot point, located at ghe middle of a 7.0 m seesaw. how far from the pivot would a person on the other side with a mass of 75 kg have to sit to balance the seesaw?
a. 2. 30 m
b. 3.00 m
c. 3.16 m
d. 3.50 m

2) a 3kg box resting on an incline of 20 degrees, when given a gentle tap, will slide at a constant speed. what is the coefficient of friction between the two surfaces?
a. .30
b. .34
c. .36
d. 1.02

Answer 1

1) Summing torque at the pivot point
64*g*2.7-75*g*x=0
2.3 m from the pivot point (a)

2) since the speed is constant, acceleration is 0
look at a FBD of the box
m*g*sin(20)-m*g*cos(20)*µk=0
solve for µk

µk=tan(20)=0.36 (c)

j