A 45.2kg block is pushed 7.00 m across a surface with coefficient of friction 0.

Question

A 45.2kg block is pushed 7.00 m across a surface with coefficient of friction 0.422 by means of a worker pushing 517N at an angle 28degrees below the horizontal.
a) What is the acceleration of the block? b)What is the work done by the worker? c) If it took 25.25 sec, what is the Power applied by the worker? A 45.2kg block is pushed 7.00 m across a surface with coefficient of friction 0.422 by means of a worker pushing 517N at an angle 28degrees below the horizontal.
a) What is the acceleration of the block? b)What is the work done by the worker? c) If it took 25.25 sec, what is the Power applied by the worker?

Explanation / Answer

A 45.2kg block is pushed 7.00 m across a surface with coefficient of friction 0.422 by means of a worker pushing 517N at an angle 28degrees below the horizontal.

1) Find the force of friction and the x component of the applied force:

Fw=mg=42.2kg(9.8m/2^2)=413.56N
Fn=413.56N
Ff=Fnµ=(413.56N)*(.42)=173.7N
So 173.7N of friction will oppose the x component of the force we'ere about to determine: To find the x component of the force: Fx=cos28(517N)=456.5N Because the Ff is opposing this motion, we'll say it's negative: SF=ma Fx-Ff=ma 456.5N-173.7N=(42.2kg)a 282.8N=(42.2kg)a 282.8N/42.2kg=6.7m/s^2 a) This is the first answer: a=6.7m/s^2

b)What is the work done by the worker?
W=Fdcos? We applied 456.5N in the x direction for 7m: W=(456.5N)(7m)=3195.5joules
c) If it took 25.25 sec, what is the Power applied by the worker? P=W/?t We already determined the work and the time is given, so plug in the information: P=3195.5joules/25.25sec=126.55watts

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.