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A mass M=15kg is pulled along a horizontal floor, with a coefficient of kinetic friction MK=0.15, for a distance of 7m. Then the mass is continued to be pulled up a frictionless incline that makes an angle theta=38 degrees with the horizontal. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of theta=38 degrees (Thus on the incline it is parallel to the surface) and has a tension T=55N.1) What is the work done by tension before the block gets to the incline?2) What is the work done by friction as the block slides on the flat horizontal surface?

A Mass M15kg Is Pulled Along A Horizontal Floor With A Coefficient Of Kinetic Friction MK015 For A Distance Of 7m Then The Mass Is Continued To Be Pulled Up A F class=

Sagot :

ANSWER:

1) 303.38 J

2) -118.93 J

STEP-BY-STEP EXPLANATION:

Given:

m = 15 kg

k = 0.15

d = 7 m

T = 55

Angle = 38°

1)

Work done by tension force is given as:

[tex]\begin{gathered} W_t=T\cdot d\cdot\cos \theta \\ \text{ replacing:} \\ W_t=55\cdot7\cdot\cos 38 \\ W_t=303.38\text{ J} \end{gathered}[/tex]

2)

For the mass , force equation in vertical direction is given as:

[tex]\begin{gathered} T\cdot\sin \theta+F_n=m\cdot g \\ \text{ replacing} \\ 55\cdot\sin 38+F_n=15\cdot9.81 \\ F_n=113.29\text{ N} \end{gathered}[/tex]

Kinetic frictional force is given as

[tex]\begin{gathered} F_k=k\cdot F_n \\ F_k=0.15\cdot113.29 \\ F_k=16.99\text{ N} \end{gathered}[/tex]

Work done by the frictional force is given as:

[tex]\begin{gathered} W_k=F_k\cdot d\cdot\cos \alpha \\ \text{ replacing} \\ W_k=16.99\cdot7\cdot\cos _{} \\ W_k=-118.93\text{ J} \end{gathered}[/tex]

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