Answer:
Taking forces along the plane
F cos θ - M g sin θ -100 = M a net of forces along the plane
F = (M a + M g * .5 + 100) / .866 solving for F
F = (80 * 1.5 + 80 * 9.8 * .5 + 100) / .866 = 707 N
F = 707 N acting along the plane
Fn = F sin θ + M g cos θ forces acting perpendicular to plane
Fn = 707 * 1/2 + 80 * 9.8 * .866 = 1030 Newtons forces normal to plane
(this would give a coefficient of friction of 100 / 1030 = .097 = Fn)
Answer:
"1155 N" is the appropriate solution.
Explanation:
Given:
Acceleration,
Forces resisting motion,
Mass,
By using Newton's second law, we get
⇒ 
Or,
⇒ 
By putting the values, we get
⇒ 
⇒ 
⇒ 
Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)
Kinetic Energy would have to be the answer. Pretty sure.
