Question

A 1 kg block in the figure above begins at rest at the top of a slope.
a. Assuming no friction, derive an expression for the speed of the block at the end of the slope if h = 0.4m and g = 9.8 m/s2.
b. If θ = 30◦, calculate the horizontal component of the velocity when the block is at the bottom end of the slope.
c. If θ = 30◦, calculate the vertical component of the velocity when the block is at the bottom end of the slope.
d. If H = 1.2 m, calculate how long the block is in freefall.
e. Calculate the distance R, the horizontal displacement of the block after it leaves the slope. f. Repeat part a, but assuming the coefficient of kinetic friction μk = 0.1.
g. Repeat part b, but assuming the coefficient of kinetic friction μk = 0.1.
h. Repeat part c, but assuming the coefficient of kinetic friction μk = 0.1.
i. Repeat part d, but assuming the coefficient of kinetic friction μk = 0.1.
j. Repeat part e, but assuming the coefficient of kinetic friction μk = 0.1.

Answer 1

a.
mgh = 1/2 mv^2
v = √2gh = √2*9.8*.4 = 2.8 m/s

b.
vx = v cos30 = 2.8 cos30 = 2.4248 m/s

c.
vy = -vsin30 = -2.8*.5 = -1.4 m/s

d.H = 1.4t+0.5*9.8t^2
1.2 = 1.4t+4.9t^2
t = 0.3722 s

e. R = vxt = 2.4248 * 0.3722 = 0.9025 m

f. mgh = 1/2 mv^2+0.1*m*g*cos30*h/sin30
9.8*0.4 = 0.5*v^2 +0.1*9.8*0.4*1.732
v = 2.775 m/s

g.
vx = v cos30 = 2.775 cos30 = 2.420327 m/s

h. vy = -vsin30 = -2.775*.5 = -1.3875 m/s
Similarly others can be solved as I have done before.