2) Suppose you have a 120.0-kg wooden crate resting on a wood floor, with coefficient of static friction 0.500 between these wood surfaces.

Question

2) Suppose you have a 120.0-kg wooden crate resting on a wood floor, with coefficient of static friction 0.500 between these wood surfaces. (a) What maximum force can you exert horizontally on the crate without moving it? (b) If you continue to exert this force once the crate starts to slip, what will its acceleration then be? The coefficient of sliding friction is known to be 0.300 for this situation

in progress 0
Thành Đạt 5 months 2021-08-24T12:55:09+00:00 1 Answers 30 views 0

Answers ( )

    1
    2021-08-24T12:56:50+00:00

    Answer:

    (a)F_{max}=588N\\(b)acceleration=1.96m/s^2

    Explanation:

    Given data

    Mass m=120.0 kg

    Coefficient of static friction u_{k}=0.500

    The coefficient of sliding friction u_{d}=0.300

    For Part (a) Maximum force

    According to Newtons second law the net force aced on the body is given by

    F_{net}=ma=F_{max}-f_{stat}=0\\so\\F_{max}=f_{stat}

    The friction force is given by

    f_{stat}=u_{stat}*N\\f_{stat}=0.5*(9.8*120)\\f_{stat}=588N

    Conclude that

    F_{max}=f_{stat}=588N

    For Part (b) Acceleration

    The acceleration due to dynamic friction is given by:

    ma=F_{max}-f_{dyn}

    The dynamic friction is given by:

    f_{dyn}=u_{dyn}*N\\f_{dyn}=0.300*(9.8*120)\\f_{dyn}=353N

    So the acceleration given by

    a=\frac{F_{max}-f_{dyn}}{m}\\ a=\frac{588N-353N}{120kg}\\ a=1.96m/s^2

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )