component of weight down slope = m g sin 30
component of weight normal to slope= m g cos 30
friction force down slope = 0.3 m g cos 30
find force up slope required for acceleration a
F = m a + m g sin 30 + 0.3 m g cos 30
= m [ a + g /2 + 0.26 g ] = m [ a + 0.76 g]
now what is a ?
d = (1/2)a t^2 if starting at constant a from dead stop
7.5 * 2 = a * 25
a = 15/25 = 0.6
so
F = 20 [ 0.6 + 0.76*9.81 ] = 161 Newtons = m g = 9.81 m
m = 16.4 kg
A block of wood of mass 20 kg is resting on a slope at 30 degrees to the
horizontal. A cord is attached to it, running parallel to the plane and over a
frictionless pulley which supports a mass (m) as shown below. Given that the
coefficient of sliding friction between the block and the slope is 0.3, calculate the
mass of m required to haul the block a distance of 7.5 metres up the slope in
5.0 seconds.
1 answer
2 answers