Respuesta :
Answer:[tex]W=-762.84\ J[/tex]
Explanation:
Given
mass of block [tex]m=16\ kg[/tex]
Force [tex]F=188\ N[/tex]
Force is applied at an angle of [tex]\theta =33.2^{\circ}[/tex]
Displacement of block [tex]s=96.1\ m[/tex]
coefficient of kinetic friction [tex]\mu _k=0.147[/tex]
Friction force acting on block [tex]f_r[/tex]
[tex]f_r=\mu _kN[/tex]
where N=Normal reaction
[tex]N=mg-F\sin \theta [/tex]
[tex]N=16\times 9.8-188\times \sin (33.2)[/tex]
[tex]N=156.8-102.94[/tex]
[tex]N=53.859\approx 54\ N[/tex]
[tex]f_r=0.147\times 54[/tex]
[tex]f_r=7.938\ N[/tex]
Work done by friction
[tex]W=f_r\cdot s[/tex]
[tex]W=7.938\times 96.1\times \cos (180)[/tex]
[tex]W=-762.84\ J[/tex]
Negative sign indicates that displacement is against the direction of force.
Answer:
Magnitude of work done W = 2.214 kW.
Explanation:
Given :
Mass of object , m = 16 kg.
Angle from horizon , [tex]\theta=33.2^o[/tex].
Force applied , F = 188 N.
Displacement of block , D = 96.1 m.
Coefficient of kinetic friction , [tex]\mu=0.147[/tex] .
Component of force, required to move the block, [tex]F\ cos\theta=188\times cos\ 33.2^o=157.31\ N.[/tex]
Also , frictional force , [tex]f=\mu N=\mu (mg)=0.147\times 16\times 9.8=23.045\ N.[/tex]
We know, work done , [tex]W = Fdcos\theta[/tex] ( Here [tex]\theta = 180^o[/tex] , because the direction of displacement is opposite the direction of frictional force .)
Putting all value in above equation :
We get , [tex]W=23.045\times 96.1\times cos\ 180^o=-2214.62 \ W=-2.214\ kW.[/tex]
Magnitude of work done W = 2.214 kW.
Hence, this is the required solution.
