wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A small block slides along a path, without friction, until the block reaches the section L=3m, which begins at height h=3m on a flat incline of angle 37o, as shown. In that section, the coefficient of kinetic friction is μk=0.50. The block passes through point A with a speed of 136m/s. Find the speed (in m/s) of the block as it passes through the point B where the friction ends. (Take g=10m/s2)
830740_1272ccb39f474238a4762e8f4e95e483.png

A
4 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
6 m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
8 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10 m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 6 m/s

According to the conservation of total energy

Mechanical energy = kinetic energy + potential energy=Constant

Let, m be the mass of a block

PE at point A

=mgh

=m×10×0=0

KE at point A=Total mechanical energy at point A

=12mv2

=12m(136)2

=68m

Let at height 3 m on inclined plane, a point C is there.

Potential energy at point C,

PEC=m×10×3=30m

Net Kinetic energy at point C

=68m30m

=38m

While the travels from C to b, There will some loss of mechanical energy due to friction which iwill given by :

E=μN×3

Normal force

N=mgcosθ

=m×10×cos(37)

=8.0m

Kinetic friction force

=μN

=0.5×8m

=4 m

Loss in total energy adue to friction

4×3=12 J

Loss of kinetic energy due to the height

=mgLsinθ

=m×10×sin(37)

=6m J

Net Kinetic energy at B

=38m12m6m

=20m J

so,12mv2=20m

v=6.3 m/s

Therefore, the velocity of the body at point B will be approximate 6 m/s.


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon