Suppose we need to find the acceleration of the sled.
60⋅cos20°−0.22N=ma60\cdot\cos20°-0.22N=ma60⋅cos20°−0.22N=ma
N+60⋅sin20°=mg→N=mg−60⋅sin20°=22⋅9.8−60⋅sin20°=195(N)N+60\cdot \sin20°=mg\to N=mg-60\cdot\sin20°=22\cdot 9.8-60\cdot\sin20°=195(N)N+60⋅sin20°=mg→N=mg−60⋅sin20°=22⋅9.8−60⋅sin20°=195(N)
a=60⋅cos20°−0.22Nm=60⋅cos20°−0.22⋅19522=0.613(m/s2)a=\frac{60\cdot\cos20°-0.22N}{m}=\frac{60\cdot\cos20°-0.22\cdot195}{22}=0.613(m/s^2)a=m60⋅cos20°−0.22N=2260⋅cos20°−0.22⋅195=0.613(m/s2). Answer
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