(a) Calculate the magnitude of the acceleration (ignore friction).
m/s2
(b) After 3.00 s, how fast will the crate be moving?
m/s
m/s2
(b) After 3.00 s, how fast will the crate be moving?
m/s
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Hello,
Draw a FBD of the crate. There is only one force on the crate, that is the weight, pointing straight down. You must break this weight into two components, one parallel to the slope and one perpendicular to the slope.
The one parallel to the slope is W * sin theta, or mg sin theta. This is the only force on the crate in the direction of movement. And sum of forces equals mass times acceleration, right?
F = ma, therefore,
a = F / m
and F = mg sin theta,
a = (mg sin theta) / m
a = g sin theta
a = 9.81 m/s^2 * sin 30
a = 4.905 m/s^2
B) Assume the initial velocity was zero. We know acceleration, initial velocity and time, and we need final velocity.....We need the equation of motion without distance in it.
Vf = Vi + a *t, and Vi = 0 m/s
Vf = a * t = 4.905 m/s^2 * 3.00 sec = 14.7 m/s
Good Luck
Draw a FBD of the crate. There is only one force on the crate, that is the weight, pointing straight down. You must break this weight into two components, one parallel to the slope and one perpendicular to the slope.
The one parallel to the slope is W * sin theta, or mg sin theta. This is the only force on the crate in the direction of movement. And sum of forces equals mass times acceleration, right?
F = ma, therefore,
a = F / m
and F = mg sin theta,
a = (mg sin theta) / m
a = g sin theta
a = 9.81 m/s^2 * sin 30
a = 4.905 m/s^2
B) Assume the initial velocity was zero. We know acceleration, initial velocity and time, and we need final velocity.....We need the equation of motion without distance in it.
Vf = Vi + a *t, and Vi = 0 m/s
Vf = a * t = 4.905 m/s^2 * 3.00 sec = 14.7 m/s
Good Luck