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Unformatted text preview: LAST NAME: STUDENT NR: PHYS 1010 6.0: CLASS TEST 2 Time: 50 minutes; Calculators & formulae provided at the end = only aid; Total = 20 points. 1) A crate (m = 1.0 × 102 kg) needs to be pulled across a smooth ﬂoor by John and Bob as shown in the ﬁgure. The friction coeﬃcients are known as µs = 0.25, µk = 0.10. The crate location at time t = 0 is shown, John pulls with FJ = 1.0 × 102 N, and Bob with FB = 2.0 × 102 N at the angles indicated. Provide answers for x(t) and y (t), i.e., for the position vector of the motion. Start with a free-body diagram (include friction!). Will the crate move? 1 2)  Derive the formula for the centripetal acceleration (acp = vr ) from the position vector describing uniform circular motion (formula sheet!), and show the direction for the acceleration vector. 2 2 3)  Calculate the earth’s linear speed in its motion around the sun starting from the law of gravity and Newton’s 2nd law. Assume dS−E = 1.5 × 1011 m, and MS = 2.0 × 1030 kg. Then calculate the length of a year from one orbit. 3 4)  When you compress a spring the force increases linearly with the displacement from equilibrium ∆x. Calculate the work associated with this compression. Do the calculation based on geometry, do not use integral calculus, i.e., start with a graph of the spring force vs displacement ∆x. By Hooke’s law F = −k ∆x, and note that ∆x can be positive or negative. FORMULA SHEET v (tf ) = v (ti ) + ti a(t) dt s(tf ) = s(ti ) + ttif v (t) dt vf = vi + a∆t sf = si + vi ∆t + 1 a∆t2 vf2 = vi2 + 2a∆s 2
tf g = 9.8 m/s2 f (t) = t f (t) = a = 0 F (t) = f (t) dt = at + C F (t) = anti-derivative = indeﬁnite integral area under the curve f (t) between limits t1 and t2 : F (t2 ) − F (t1 ) x2 + px + q = 0 factored by: x1,2 = − p ± p4 − q 2 ˆ + sin ω t ˆ); (t) = uniform circular m. (t) = R(cos ω t i r jv exp = exp; sin = cos; cos = − sin. m = Fnet ; a FG =
Gm1 m2 ; r2 d [f (g (x))] dx df dt df dt = 1 F (t) = f (t) dt = t2 2 +C 2 = d r dt df dg ; dg dx = ...; (t) = a
d v dt = .... (f g ) = f g + f g
2 g= GME 2; RE fs ≤ µs n; fk = µk n; fr = µr n; µr << µk < µs . FH = −k ∆x = −k (x − x0 ). 2 Fd ∼ − ; linear: Fd = dv ; quadratic: Fd = 0.5ρAv ; A = cross sectional area v W = F ∆x = F (∆r) cos θ For F (x) the work is given as area under the Fx vs x curve. 4 RE = 6370 km; G = 6.67 × 10−11 Nm2 ; ME = 6.0 × 1024 kg kg ...
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