A block of mass M = 10 kg is acted on by an applied force F_1 directed at 45 deg
ID: 1362552 • Letter: A
Question
A block of mass M = 10 kg is acted on by an applied force F_1 directed at 45 degree above the horizontal as shown. The block slides on a surface with mu_k= 0.20 and. starting from rest, achieves a speed of 4 m/s after sliding a distance d = 2 m. What is the work done by the net force on the block? What is the net force on the block? What is the magnitude of F_1? A 10-kg block (m_1) attached by a rope to a 5-kg block (m_2) is pulled by another rope at constant speed of 2 m/s up a frictionless incline making an angle of 15 degree with the horizontal. What is the tension T_1? What is the tension T_2?Explanation / Answer
from the work energy theorem , work done is equal to change in KE
A)
work done = change in KE
W = 0.5*M*(vf^2 - vi^2)
W = 0.5*10*(4^2-0) = 80 J
(B)
net work done = Fnet *d
80 = Fnet * 2
Fnet = 40 N
(C)
from the figure
along perepndicular to the surface
F1*sin45 + N = Mg
N = Mg - F1*sin45
frictional force fk = uk*N = uk*(Mg - F1*sin45)
net force Fnet = F1*cos45 - fk
40 = F1*cos45 - 0.2*(10*9.8 - F1*sin45)
F1 = 70.2 N
+++++++++++++
2)
here the apeed is constant. so, acceleration a = 0
for m2
T2 - m2*g*sintheta = 0
T2 = m2*g*sintheta
for m1
T1 - m1*g*sintheta - T2 = 0
T1 - m1*g*sintheta -m2*g*sintheta = 0
T1 = (m1+m2)*g*sin15
T1 = (10+5)*9.8sin15 = 38 N
T2 = m2*g*sin15 = 5*9.8*sin15 = 12.7 N
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.