Answer:
Explanation:
We apply Newton's second law at the crate :
∑F = 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:
m=90kg : crate mass
F= 282 N
μk =0.351 :coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Crate weight (W)
W= m*g
W= 90kg*9.8 m/s²
W= 882 N
Friction force : Ff
Ff= μk*N Formula (2)
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W = 0
N = W
N = 882 N
We replace the data in the formula (2)
Ff= μk*N = 0.351* 882 N
Ff= 309.58 N
We apply the formula (1) in x direction:
∑Fx = m*ax , ax=0
282 N - 309.58 N = 90*a
a= (282 N - 309.58 N ) / (90)
a= - 0.306 m/s²
Kinematics of the crate
Because the crate moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d Formula (3)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = 0.850 m/s
d = 0.75 m
a= - 0.306 m/s²
We replace the data in the formula (3)
vf²=(0.850)²+(2)( - 0.306 )(0.75 )
The momentum of two or more objects during collisions is not lost nor gained
Answer:
3.13 m/s
Explanation:
From the question,
Since the flea spring started from rest,
Ek = W................... Equation 1
Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.
But,
Ek = 1/2mv²............ Equation 2
Where m = mass of the flea spring, v = flea's speed when it leaves the ground.
substitute equation 2 into equation 1
1/2mv² = W.................... Equation 3
make v the subject of the equation
v = √(2W/m)................. Equation 4
Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg
Substitute into equation 4
v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]
v = 7.2/2.3
v = 3.13 m/s
Hence the flea's speed when it leaves the ground = 3.13 m/s
Answer:
100 Joule
Explanation:
Amount of heat in agiven body is given by Q = m•C•ΔT
where m is the mass of the body
c is the specific heat capacity of body. It is the amount of heat stored in 1 unit weight of body which raises raises the temperature of body by 1 unit of temperature.
ΔT is the change in the temperature of body
___________________________________________
coming back to problem
m = 5g
C = 2J/gC
since, it is given that temperature of body increases by 10 degrees, thus
ΔT = 10 degrees
Using the formula for heat as given
Q = m•C•ΔT
Q = 5* 2 * 10 Joule= 100 Joule
Thus, 100 joule heat must be added to a 5g substance with a specific heat of 2 J/gC to raise its temperature go up by 10 degrees.
