The force acting in the front direction is the 130N.
The frictional force is acting backwards 30N.
1) The net force is 130N - 30N = 100N
2) s = ut + (1/2)at^2 u = 0, Start from rest, s = 25m t =5.
25 = 0*5 + (1/2)* a * 5^2.
25 = 0 + 25/2 * a.
25 = (25/2)a. Divide 25 from both sides.
1 = (1/2)* a. Cross multiply.
2 = a.
a = 2 m/s^2.
3) Mass of the box
Net Force, F = ma
100 = m*2. Divide both sides by 2.
100/2 = m
50 = m.
m = 50 kg.
4) Final velocity , v = u + at.
v = 0 + 2*5 = 10 m/s.
Kinetic Energy, K = (1/2) * mv^2.
= 1/2 * 50 * 10 * 10.
= 2500 J.
Answer:
μk = 0.26885
Explanation:
Conceptual analysis
We apply Newton's second law:
∑Fx = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
a= -0.9 m/s²,
g = 9.81 m/s² : acceleration due to gravity
W= 75 N : Block weight
W= m*g
m = W/g = 75/9.8= 7.65 kg : Block mass
Friction force : Ff
Ff= μk*N
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W-25 = 0
N = 75
+25
N= 100N
∑Fx = m*ax
20-Ff= m*ax
20-μk*100
= 7.65*(-0.90 )
20+7.65*(0.90) = μk*100
μk = ( 20+7.65*(0.90)) / (100)
μk = 0.26885
Answer:3.68 kg
Explanation:
Given
mass
distance traveled by is 50 cm in 1 s
using
Suppose T is the tension in string and is the mass of other body
For
for other body
Equating value of Tension
Answer:
0.7432m²
1152in²
Explanation:
a) Usng the conversion
1ft² = 0.0929m²
8ft² = y
y = 8 × 0.0929
y = 0.7432m²
Hence 8ft² in m² is 0.7432m²
For ft² to in²
1ft² = 144in²
8ft² = x
X =8×144
x = 1152in²
Hence 8ft² expressed in in² is 1152in²
The correct answer is 3×10^8 m/s