When block is pushed upwards along the inclined plane
the net force applied on the block will be given as
[tex]F_{net} = mg sin\theta + \mu_k mg cos\theta[/tex]
here we know that
m = 75 kg
[tex]\theta = 8.5 degree[/tex]
[tex]\mu_k = 0.16[/tex]
now plug in all values into this
[tex]F_{net} = 75\times 9.8 sin8.5 + 0.16 \times 75\times 9.8 cos8.5[/tex]
[tex]F = 225 N[/tex]
now for finding the power is given as
[tex]P = Fv[/tex]
[tex]0.87 \times 10^3 = 225 \time v[/tex]
[tex]v = \frac{870}{225} = 3.87 m/s[/tex]
The maximum constant speed of block under the power applied by the student is 3.86 m/s.
Given data:
The magnitude of power is, P = 0.87 kW = 870 W.
The mass of block is, m = 75 kg.
Angle of inclination with horizontal is, [tex]\theta = 8.5^\circ[/tex].
The coefficient of kinetic friction between the block and incline is, [tex]\mu = 0.16[/tex].
he expression for the net force on the block at incline is,
[tex]F=mgsin\theta + \mu mgcos \theta\\F=(75 \times 9.8 \times sin8.5) + (0.16\times 75 \times cos 8.5)\\F = 225 \;\rm N[/tex]
The expression for the power is,
[tex]P = F \times v[/tex]
Here, v is the maximum constant speed of block under the power applied by the student.
Solving as,
[tex]870 = 225 \times v\\v = \dfrac{870}{225}\\v = 3.86 \;\rm m/s[/tex]
Thus, the maximum constant speed of block under the power applied by the student is 3.86 m/s.
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