Answer:
F = m g sin theta force accelerating block
m a = m g sin theta
a = 9.8 sin 24 = 3.99 m/sec^2
Answer:
2420 J
Explanation:
From the question given above, the following data were obtained:
Force (F) = 22.9 N
Angle (θ) = 35°
Distance (d) = 129 m
Workdone (Wd) =?
The work done can be obtained by using the following formula:
Wd = Fd × Cos θ
Wd = 22.9 × 129 × Cos 35
Wd = 22.9 × 129 × 0.8192
Wd ≈ 2420 J
Thus, the workdone is 2420 J.
Answer:
Magnitude of the force is 2601.9 N
Explanation:
m = 450 kg
coefficient of static friction μs = 0.73
coefficient of kinetic friction is μk = 0.59
The force required to start crate moving is 
.
but once crate starts moving the force of friction is reduced .
Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie .
Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.
Magnitude of the force
The speed is changing its direction all the time. There
is an acceleration which changes the direction of the speed – that is called
centripetal acceleration. Only uniform linear motions are considered to have no
acceleration.
This is the general formula for acceleration
a = dv/dt
When calculating dv, you should keep in mind the change
in the velocity vector’s direction. You can easily see in a graph that with dt
tending to 0 (so the length of the arc covered is also tending to 0), the difference
between vectors Vf and V0 has a direction which is perpendicular to velocity
(the shorter the arc, the closest the angle is to 90 degrees).
There is a formula (which can be deducted from the
previous formula) which allows you to calculate the acceleration:
a = v^2/r
Let’s talk about the units:
v is in m/s
r is in m
so v^2/r
is in (m/s)^2/m = (m^2/s^2)/m = m/s^2
which is the same unit as dv/dt:
dv/dt = (m/s)/s= m/s^2
