Respuesta :
Answer:
(a).The work done by the constant force is 239.5 J.
(b). The coefficient of friction is 0.12
Explanation:
Given that,
Force = 30 N
Mass of block = 16 kg
Angle = 37°
Distance = 10 m
Velocity = 3 m/s
(a). We need to calculate the work done by the constant force
Using formula of work done
[tex]W=F\cos\theta\cdot d[/tex]
Put the value into the formula
[tex]W=30\cos37\times10[/tex]
[tex]W=239.59\ J[/tex]
(b). If at the start of the motion the block was at rest and after 10 meters it has velocity of magnitude |v| = 3 m/s,
We need to calculate the acceleration
Using equation of motion
[tex]v^2-u^2 =2as[/tex]
[tex]a=\dfrac{v^2-u^2}{2s}[/tex]
Put the value into the formula
[tex] a=\dfrac{9-0}{2\times10}[/tex]
[tex]a=0.45\ m/s^2[/tex]
We need to calculate the coefficient of friction
Using formula of frictional force
[tex]f=\mu\times N[/tex]...(I)
Here, normal force is
[tex]N=mg-F\sin\theta[/tex]
Net force is
[tex]F=F\cos\theta-f[/tex]
[tex]f=F\cos\theta-F[/tex]
Put the value of N in equation (I)
[tex]f=\mu(mg-F\sin\theta)[/tex]
[tex]F\cos\theta-F=\mu(mg-F\sin\theta)[/tex]
[tex]F=F\cos\theta-\mu(mg-F\sin\theta)[/tex]
[tex]ma=F\cos\theta-\mu(mg-F\sin\theta)[/tex]
[tex]\mu=\dfrac{F\cos\theta-ma}{mg-F\sin\theta}[/tex]
Put the value into the formula
[tex]\mu=\dfrac{30\cos37-16\times0.45}{16\times9.8-30\sin37}[/tex]
[tex]\mu=0.12[/tex]
Hence, (a).The work done by the constant force is 239.5 J.
(b). The coefficient of friction is 0.12
