Answer:
if you are asking for static friction the answer is <u>372N</u>
Explanation
friction static=normal force x coefficient of static friction
and normal force is equal to weight of a body
so static friction = 600N x 0.62= 372N
The force of earth gravity on the space craft is 1.19 N
\begin{array}{c}
F = \frac{{GMm}}{{{{\left( {r + h} \right)}^2}}}\\
= \frac{{6.67 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}} \times 1050}}{{{{\left( {\left( {6400 + 12800} \right) \times {{10}^3}} \right)}^2}}}\\
= 1.139\;{\rm{N}}
\end{array}$
Hence, the force of gravity on the space craft is 1.19 N.
Answer:
5.85 m/s²
Explanation:
Mass of pulley (M) = 6.8 kg, mass of bucket (m) = 1.5 kg, radius of pulley (r) = 0.816 m, acceleration due to gravity (g) = 10 m/s².
Let the tension exerted on the pulley be T and the angular acceleration be α,
The angular acceleration (α) = a / r; where a is the linear acceleration.
mg - T = ma
T = m(g - a)
Also, for the disk:
Tr = Iα
I = moment of inertia = 0.5Mr²
m(g - a)r = 0.5Mr²(a/r)
m(g - a)r = 0.5Mra
2m(g - a) = Ma
2mg - 2ma = Ma
Ma + 2ma = 2mg
a(M + 2m) = 2mg
a = 2mg / (M + 2m)
Substituting:
a = 2(3.1 kg)(10 m/s²) / (4.4 + 2(3.1))
a = 5.85 m/s²
Answer:
Part a)
Work done is given as
W = -121.5 J
Part b)
Potential energy of the system is given as
Explanation:
As we know that work done by variable force is given as
here we know that
so we have
so we have
here we displace it from x = 2 to x = 5
so we will have
Part b)
As we know that change in Potential energy = work done
Answer:
432 J
Explanation:
When moving linearly:
Kinetic Energy = (1/2)mV^2
So here you have:
KE=(1/2)(6)(12^2)=(1/2)(6)(144)=432
The unit for energy is Joules (J), so your answer would be 432 J.