phy101 assignment no 2 Fall November 2011 solution

[Vuhelper]

Question # 1 Marks = 12

A worker pushes 25Kg crate a distance of 5m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic frication between the crate and the floor is 0.23.

(a) What magnitude of force must the worker apply?

(b) How much work is done on the crate by this force?

(c) How much work is done on the crate by frication?

(d) How much work is done on the crate by normal force?

(e) How much work is done on the crate by the gravity?

(f) What is the total work done on the crate?



Question # 2 Marks = 8

The flywheel of a prototype car engine is under test. The angular position θ of the flywheel is given by θ = (3.0rad/s3)t3 and the diameter of the flywheel is 36cm.



(a)Find the distance that a particle on the rim moves during that time interval.

(b) Find the angle θ, in radians and in degree, at times t1 = 3.0s and t2 = 6.0s.

(c) Find the average angular velocity, in rad/s and in rev/min, between t1 = 3s and t2 = 6s.

(d) Find the instantaneous angular velocity at time t1 = 3.0s and t2 = 6.0s.

[Vuhelper]

Question No. 1:
A worker pushes 25kg crate a distance of 5m along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.23.
a) What magnitude of force must the worker apply?
b) How much work is done on the crate by this force?
c) How much work is done on the crate by friction?
d) How much work is done on the crate by normal force?
e) How much work is done on the crate by the gravity?
f) What is the total work done on the crate?

Solution:
Mass = 25kg
Distance = 5m
Coefficient of friction =0.23N
W = mg
W = 25 × 9.8
W = 245 N
Fc = 245N
Fp =
Fp = 0N
Fv = 245cos 0
Fv = 245N
Ff = Fv
Ff =
Ff = 56.35 N

a) What magnitude of force must the worker apply?

Fap = Fp + Ff
Fap = 0 + 56.35
Fap = 56.35N

b) How much work is done on the crate by this force?

W = Applied Force × Distance
W = Fap × d
W = 56.35 × 5
W = 281.75 J

c) How much work is done on the crate by friction?

W = Force of Friction × Distance
W = Ff × d
W = 56.35 × 5
W = 281.75 J

d) How much work is done on the crate by normal force?

W = Fv × h
W = 245 × 0
W = 0 J

e) How much work done on the crate by the gravity?

W = Fc × h
W = 245 × 0
W = 0 J

f) What is the total work done on the crate?

W = 281.75 + 281.75 + 0 + 0
W = 563.5 J

Question No. 2:
The flywheel of a prototype car engine is under test. The angular position of the flywheel is given by = (3.0rad/s3)t3 and the diameter of the flywheel is 36 cm.

a) Find the distance that a particle on the rim moves during that time interval.
b) Find the angle , in radians and in degree, at times t1 = 3.0s and t2 = 6.0s.
c) Find the average angular velocity, in rad/s and in rev/min, between t1 = 3s and t2 = 6s
d) Find the instantaneous angular velocity at time t1 = 3.0s and t2 = 6.0s.

Solution:

a) Find the distance that a particle on the rim moves during that time interval.


Diameter = d = 36cm
= 3.0 rad/sec
Radius = r = 36/2
R = 18 cm
Distance = S =
S =
S = 54 cm
b) Find the angle , in radians and in degree, at times t1 = 3.0s and t2 = 6.0s.
Radian made in one second = 3 rad
Radian made at t1 = 3 × 3
= 9 rad
Radian made at t2 = 3 × 6
= 18 rad
c) Find the average angular velocity, in rad/s and in rev/min, between t1 = 3s and t2= 6s
Average velocity at t2 =


Average velocity at t1 =



d) Find the instantaneous angular velocity at time t1 = 3.0s and t2 = 6.0s.
Instantaneous angular velocity at t1

Instantaneous angular velocity at t2