1. A 20kg block is resting in a horizontal surface. Calculate the force required to move it if the coefficient of static and dynamic friction are 0.75 and 0.6 respectively.

Solution:

F= (COF) N

N=W=mg= (20kg) (9.81m/s2)

F= 0.75 (20) (9.81) ; use static COF if the body is at rest

**F= 147.15 N (B)**

2. A 100kg object is sliding over a surface with static and dynamic coefficient of friction values of 0.5 and 0.3 respectively. If the initial velocity of the object is 10m/s, how long will it travel before it completely stops?

Solution:

FBD

Assuming dynamic equilibrium:

Summation of horizontal force:

REF-Fr=0

REF=Fr

REF= (W/g) (a)

Fr= (COF) (N)

Summation of Vertical forces:

N=W

(W/g)(a)=(COF)(N)

a/g=COF

a= (COF) (g)

Note: Use dynamic COF value since the body is already in motion.

a=0.3 (9.81) = 2.943 m/s2

a=-2.943 (negative, because the body is decelerating)

Using the rectilinear motion equation:

V^2= v0^2+2aS

S= (V2-v02)/ (2a)

Since the final velocity is zero

S= (0-10) ^2/ (2) (-2.943)

S=16.989 m (A)

3. A 1kg ball is tied at the end of a 1.5m long rope and then rotated at the speed of 100 RPM. Calculate the tension on the rope.

F=ma;

a=v^2/r (normal acceleration for revolving bodies)

v= (pi)(D)(N)/60=2pirN/60

v=2(3.1416)(1.5)(100/60)

v=15.708 m/s

a= (15.708)^2/1.5

a=164.5 m/s2

F=1(164.5)

**F=164.5 N (D)**

4. A bullet is fired vertically from a gun situated 2 m above the ground. If the initial speed of the bullet is 100m/s, how high will it travel before it starts falling? Assume negligible air resistance.

V^2=v0^2+2as; a=-g (decelerating at the value of freely falling acceleration)

V=0; since at the highest point, the bullet will stop and starts falling down.

0=100^2+2 (-9.81)(S)

S=509.7 m+2m

**S=511.7 m (D)**

5. A missile is fired from North Korea and is targeted towards South Korea. If the distance between the two countries is 100km and the firing angle is 45 degrees, at what velocity should the missile be fired to reach its target? Assume negligible air resistance.

From projectile derived equations:

Range=v0^2 (sin2X)/g

100000m= (v0^2) (sin(90)/(9.81m/s2)

**V0=990.45 m/s (C)**

6. Calculate the polar moment of inertia of a 30 cm diameter circular board.

J=(pi)D^4/32

J=(pi) (30 cm) ^4/32

J=(3.1416)25312.5 cm^4

**J=79521.6 cm^4**

7. A 20mm thick, 50 cm in diameter wheel made of cast iron (density: 6.0kg/cu m) starts rotating from rest and in 1 minute, attained an angular speed of 200 RPM. Calculate the angular acceleration in rad/s2?

α= (ω- ω0)/t

ω0=0 (from rest)

ω=200 rev/min (1min/60s) (2pi rad/rev)

ω = 20.944 rad/s

α = (20.944 rad/s)/ (60s)

**α =0.349 rad/s2 (B)**

8. A 2kg ball starts rolling down from the top of a 100m-high hill towards the plains. Calculate the final speed of the ball when it reaches the foot of the hill. Assume negligible friction.

V^2=v0^2+2as; a=g=9.81m/s2

V^2=0+2(9.81) (100)

**V=44.294 m/s (D)**

9. A 500kg car is moving around a 500 m radius road with 5 degrees banking. If the minimum friction between the tires and the road is 0.2 during rainy days, calculate the maximum speed limit.

Tan (X) =v^2/gR; for speed limit computation, the assumption is NO FRICTION (for safety factor)

Tan (5) =v^2/ ((9.81) (500))

V= 20.715 m/s

**V=74 kph (C)**

10. A 2kg object with a velocity of 10m/s hits a 3kg object at rest, sticks with it and move together in the same direction. Find the resulting velocity.

Conservation of momentum

m1v1+m2v2= (m1+m2) (v)

v= (m1V1+m2v2)/ (m1+m2)

v= (2kg (10m/s) +3kg (0))/ (2kg+3kg)

**v = 4m/s (C)**

**For any question/correction/suggestion please comment below or email me at admin@tupmechanical.com.**

My apologies if the solution is not easy to follow. It's hard to write computation solution in a webpage :-)

gud day sa inyo..pde po b tau magshare ng mga matrials na magtatake ng board s september?

ReplyDeleteCorrection: Item no. 6. Polar Moment of inertia for cicular section, J = (ΠD^4)/32. So for a 30cm dia, J = (Π*30^4)/32 = 79521.56cm^4.

ReplyDeleteThanks, silly mistake it is. I forgot the pi.

ReplyDeleteAnyway, thanks a lot. I can only rely on my readers to check my errors. Please continue doing this. God bless!