ball player is running to second base at 5.03 m/s. When he is 4.80 m from the 28) plate he goes into a slide. The coefficient of kinetic friction between the player and the ground is 0.180, and the coefficient of static friction is 3.14. What is his speed when he reaches the plate? A) 1.96 m/s C) 2.89 m/s B) 4.47 m/s D) He stops before reaching the plate. 29 A heat engine with an efficiency of 30% performs 2500 J of work. How much heat is 2 discharged to the lower temperature reservoir? A) 1400 J B) 750 J C) 7100 J D) 5800 J
Solution
initial speed=5.03 m/s
distance to be covered=4.8 m
deceleration=friction coefficient*acceleration due to gravity
=0.18*9.8=1.764 m/s^2
let speed when he reaches the plate be v
then using the formula:
final speed^2-initial speed^2=2*acceleration*distance
==>v^2-5.03^2=2*(-1.764)*4.8
==>v=2.8925 m/s
hence option C is correct.
Q29.
efficiency=work done/heat input
==>0.3=2500/het input
==>heat input=2500/0.3=8333.3 J
then heat discharged to lower temperature reservoir=heat input-work done
=5833.3 J
hence option D is correct.
.
Disha NEET Physics Guide for classes 11 and 12.pdf
ball player is running to second base at 5-03 m-s- When he is 4-80 m f.docx
1. ball player is running to second base at 5.03 m/s. When he is 4.80 m from the 28) plate he goes
into a slide. The coefficient of kinetic friction between the player and the ground is 0.180, and
the coefficient of static friction is 3.14. What is his speed when he reaches the plate? A) 1.96 m/s
C) 2.89 m/s B) 4.47 m/s D) He stops before reaching the plate. 29 A heat engine with an
efficiency of 30% performs 2500 J of work. How much heat is 2 discharged to the lower
temperature reservoir? A) 1400 J B) 750 J C) 7100 J D) 5800 J
Solution
initial speed=5.03 m/s
distance to be covered=4.8 m
deceleration=friction coefficient*acceleration due to gravity
=0.18*9.8=1.764 m/s^2
let speed when he reaches the plate be v
then using the formula:
final speed^2-initial speed^2=2*acceleration*distance
==>v^2-5.03^2=2*(-1.764)*4.8
==>v=2.8925 m/s
hence option C is correct.
Q29.
efficiency=work done/heat input
==>0.3=2500/het input
==>heat input=2500/0.3=8333.3 J
2. then heat discharged to lower temperature reservoir=heat input-work done
=5833.3 J
hence option D is correct.
